Add properties of the integral of a step function.
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@ -51,10 +51,10 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}.
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\end{definition}
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\section{\partial{Integral of Step Function}}%
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\section{\partial{Integral of a Step Function}}%
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\label{sec:def-integral-step-function}
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Lset $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let
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Let $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let
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$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{sec:def-partition} of $[a, b]$
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such that $s$ is constant on the open subintervals of $P$.
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Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval,
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@ -576,7 +576,7 @@ For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
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\end{axiom}
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\section{\defined{Additive Property}}%
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\label{sec:additive-property}
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\label{sec:area-additive-property}
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If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
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$\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$.
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@ -589,7 +589,7 @@ If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
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\end{axiom}
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\section{\defined{Difference Property}}%
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\label{sec:difference-property}
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\label{sec:area-difference-property}
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If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
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$\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$.
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@ -602,7 +602,7 @@ If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
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\end{axiom}
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\section{\defined{Invariance Under Congruence}}%
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\label{sec:invariance-under-congruence}
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\label{sec:area-invariance-under-congruence}
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If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
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also in $\mathscr{M}$ and we have $a(S) = a(T)$.
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@ -615,7 +615,7 @@ If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
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\end{axiom}
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\section{\defined{Choice of Scale}}%
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\label{sec:choice-scale}
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\label{sec:area-choice-scale}
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Every rectangle $R$ is in $\mathscr{M}$.
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If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
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@ -628,7 +628,7 @@ If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
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\end{axiom}
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\section{\partial{Exhaustion Property}}%
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\label{sec:exhaustion-property}
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\label{sec:area-exhaustion-property}
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Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so
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that
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@ -664,8 +664,8 @@ A set consisting of a single point.
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Let $S$ be a set consisting of a single point.
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By definition of a point, $S$ is a rectangle in which all vertices coincide.
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By \nameref{sec:choice-scale}, $S$ is measurable with area its width times
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its height.
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By \nameref{sec:area-choice-scale}, $S$ is measurable with area its width
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times its height.
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The width and height of $S$ is trivially zero.
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Therefore $a(S) = (0)(0) = 0$.
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@ -698,7 +698,7 @@ A set consisting of a finite number of points in a plane.
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By construction, $S_{k+1} = S_k \cup T$.
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By the induction hypothesis, $S_k$ is measurable with area $0$.
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By \nameref{sub:exercise-1.7.1a}, $T$ is measurable with area $0$.
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By the \nameref{sec:additive-property}, $S_k \cup T$ is
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By the \nameref{sec:area-additive-property}, $S_k \cup T$ is
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measurable, $S_k \cap T$ is measurable, and
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\begin{align}
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a(S_{k+1})
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@ -746,8 +746,8 @@ The union of a finite collection of line segments in a plane.
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Consider a set $S$ consisting of a single line segment in a plane.
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By definition of a line segment, $S$ is a rectangle in which one side has
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dimension $0$.
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By \nameref{sec:choice-scale}, $S$ is measurable with area its width $w$
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times its height $h$.
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By \nameref{sec:area-choice-scale}, $S$ is measurable with area its width
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$w$ times its height $h$.
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Therefore $a(S) = wh = 0$.
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Thus $P(1)$ holds.
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@ -761,7 +761,7 @@ The union of a finite collection of line segments in a plane.
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By construction, $S_{k+1} = S_k \cup T$.
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By the induction hypothesis, $S_k$ is measurable with area $0$.
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By the base case, $T$ is measurable with area $0$.
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By the \nameref{sec:additive-property}, $S_k \cup T$ is measurable,
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By the \nameref{sec:area-additive-property}, $S_k \cup T$ is measurable,
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$S_k \cap T$ is measurable, and
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\begin{align}
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a(S_{k+1})
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@ -823,10 +823,10 @@ Prove that every triangular region is measurable and that its area is one half
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\centering
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\end{figure}
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By \nameref{sec:choice-scale}, both $R$ and $S$ are measurable.
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By \nameref{sec:area-choice-scale}, both $R$ and $S$ are measurable.
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By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$.
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By the \nameref{sec:additive-property}, $R \cup S$ and $R \cap S$ are both
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measurable.
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By the \nameref{sec:area-additive-property}, $R \cup S$ and $R \cap S$ are
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both measurable.
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$a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that
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$R$'s construction implies identity $a(R) = 2a(T)$.
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Therefore
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@ -839,8 +839,8 @@ Prove that every triangular region is measurable and that its area is one half
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& = ab + ca\sin{\theta} - ca\sin{\theta} - a(T).
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\end{align*}
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Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$
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By \nameref{sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our
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proof.
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By \nameref{sec:area-invariance-under-congruence}, $a(T') = a(T)$, concluding
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our proof.
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\end{proof}
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@ -866,10 +866,10 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
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Suppose $S$ is a right trapezoid.
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Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and
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height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$.
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By \nameref{sec:choice-scale}, $R$ is measurable.
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By \nameref{sec:area-choice-scale}, $R$ is measurable.
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By \nameref{sec:exercise-1.7.2}, $T$ is measurable.
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By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both
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measurable and
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By the \nameref{sec:area-additive-property}, $R \cup T$ and $R \cap T$ are
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both measurable and
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\begin{align*}
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a(S)
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& = a(R \cup T) \\
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@ -888,8 +888,8 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
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Then $R$ has longer base edge of length $b_2 - c$.
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By \nameref{sec:exercise-1.7.2}, $T$ is measurable.
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By Case 1, $R$ is measurable.
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By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both
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measurable and
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By the \nameref{sec:area-additive-property}, $R \cup T$ and $R \cap T$ are
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both measurable and
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\begin{align*}
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a(S)
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& = a(T) + a(R) - a(R \cap T) \\
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@ -906,7 +906,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
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Let $c$ denote the length of base $T$.
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Reflect $T$ vertically to form another right triangle, say $T'$.
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Then $T' \cup R$ is an acute trapezoid.
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By \nameref{sec:invariance-under-congruence},
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By \nameref{sec:area-invariance-under-congruence},
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\begin{equation}
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\label{sec:exercise-1.7.3-eq1}
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\tag{3.1}
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@ -933,7 +933,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
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Let $c$ denote the length of base $T$.
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Reflect $T$ vertically to form another right triangle, say $T'$.
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Then $T' \cup R$ is an acute trapezoid.
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By \nameref{sec:invariance-under-congruence},
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By \nameref{sec:area-invariance-under-congruence},
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\begin{equation}
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\label{sec:exercise-1.7.3-eq2}
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\tag{3.2}
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@ -970,7 +970,7 @@ Prove that the formula is valid for rectangles with sides parallel to the
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We assume $P$ has three non-collinear points, ruling out any instances of
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points or line segments.
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By \nameref{sec:choice-scale}, $P$ is measurable with area $a(P) = wh$.
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By \nameref{sec:area-choice-scale}, $P$ is measurable with area $a(P) = wh$.
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By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and
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$B = 2(w + h)$ lattice points on its boundary.
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The following shows the lattice point area formula is in agreement with
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@ -1082,7 +1082,7 @@ Use induction on the number of edges to construct a proof for general polygons.
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& = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
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& = a(S) + a(T). & \text{base case}
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\end{align*}
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By the \nameref{sec:additive-property}, $S \cup T$ is measurable,
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By the \nameref{sec:area-additive-property}, $S \cup T$ is measurable,
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$S \cap T$ is measurable, and
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\begin{align*}
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a(P)
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@ -1681,4 +1681,227 @@ This property is described by saying that every step function is a linear
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\end{proof}
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\chapter{Properties of the Integral of a Step Function}%
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\label{chap:properties-integral-step-function}
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\section{\partial{Additive Property}}%
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\label{sec:step-additive-property}
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Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$.
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Then
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$$\int_a^b \left[ s(x) + t(x) \right] \mathop{dx} =
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\int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.$$
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\begin{proof}
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Let $s$ and $t$ be step functions on closed interval $[a, b]$.
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
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Likewise, there exists a partition $P_t$ such that $t$ is constant on each
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open subinterval of $P_t$.
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Therefore $s + t$ is a step function with step partition
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$$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\},$$ the common refinement of
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$P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
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$s$ and $t$ remain constant on every open subinterval of $P$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
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Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b \left[ s(x) + t(x) \right] \mathop{dx}
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& = \sum_{k=1}^n (s_k + t_k) \cdot (x_k - x_{k-1}) \\
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& = \sum_{k=1}^n \left[ s_k \cdot (x_k - x_{k-1}) +
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t_k \cdot (x_k - x_{k-1}) \right] \\
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& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) +
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\sum_{k=1}^n t_k \cdot (x_k - x_{k-1}) \\
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& = \int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.
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\end{align*}
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\end{proof}
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\section{\partial{Homogeneous Property}}%
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\label{sec:step-homogeneous-property}
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Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$.
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For every real number $c$, we have
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$$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$
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\begin{proof}
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Let $s$ be a step function on closed interval $[a, b]$.
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
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subinterval of $P$.
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Then $c \cdot s$ is a step function with step partition $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b c \cdot s(x) \mathop{dx}
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& = \sum_{k=1}^n (c \cdot s(x)) \cdot (x_k - x_{k-1}) \\
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& = \sum_{k=1}^n c \cdot (s(x) \cdot (x_k - x_{k-1})) \\
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& = c \sum_{k=1}^n s(x) \cdot (x_k - x_{k-1}) \\
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& = c \int_a^b s(x) \mathop{dx}.
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\end{align*}
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\end{proof}
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\section{\partial{Linearity Property}}%
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\label{sec:step-linearity-property}
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Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$.
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For every real $c_1$ and $c_2$, we have
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$$\int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} =
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c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x) \mathop{dx}.$$
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\begin{proof}
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Let $s$ and $t$ be step functions on closed interval $[a, b]$.
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Let $c_1$ and $c_2$ be real numbers.
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Then $c_1 \cdot s$ and $c_2 \cdot t$ are step functions.
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Then
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\begin{align*}
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& \int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} \\
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& = \int_a^b c_1s(x) \mathop{dx} + \int_a^b c_2t(x) \mathop{dx}
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& \textref{sec:step-additive-property} \\
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& = c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x). \mathop{dx}
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& \textref{sec:step-homogeneous-property}
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\end{align*}
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\end{proof}
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\section{\partial{Comparison Theorem}}%
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\label{sec:step-comparison-theorem}
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Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$.
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If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
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$$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$
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\begin{proof}
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Let $s$ and $t$ be step functions on closed interval $[a, b]$.
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
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Likewise, there exists a partition $P_t$ such that $t$ is constant on each
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open subinterval of $P_t$.
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Let $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\}$$ be the common refinement
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of $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
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By construction, $P$ is a step partition for both $s$ and $t$.
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Thus $s$ and $t$ remain constant on every open subinterval of $P$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
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Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\
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& < \sum_{k=1}^n t_k \cdot (x_k - x_{k-1}) & \text{by hypothesis} \\
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& = \int_a^b t(x) \mathop{dx}.
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\end{align*}
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\end{proof}
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\section{\partial{Additivity With Respect to the Interval of Integration}}%
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\label{sec:step-additivity-with-respect-interval-integration}
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Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$.
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Then
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$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} =
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\int_a^b s(x) \mathop{dx} \quad\text{if}\quad a < c < b.$$
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\todo{This holds for any arrangement of values $a$, $b$, and $c$.}
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\begin{proof}
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Let $s$ be a step function on closed interval $[a, b]$.
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P$ such that $s$ is constant on each open subinterval of $P$.
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Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $c$
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as a subdivision point.
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Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
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that $x_i = c$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $Q$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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& = \sum_{k=1}^n s_k \cdot (x_k - x_{k - 1}) \\
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& = \sum_{k=1}^i s_k \cdot (x_k - x_{k - 1}) +
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\sum_{k=i+1}^n s_k \cdot (x_k - x_{k - 1}) \\
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& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx}.
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\end{align*}
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\end{proof}
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\section{\partial{Invariance Under Translation}}%
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\label{sec:step-invariance-under-translation}
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Let $s$ be a step function on closed interval $[a, b]$.
|
||||
Then
|
||||
$$\int_a^b s(x) \mathop{dx} =
|
||||
\int_{a+c}^{b+x} s(x - c) \mathop{dx} \quad\text{for every real } c.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $s$ be a step function on closed interval $[a, b]$.
|
||||
By definition of a step function, there exists a \nameref{sec:def-partition}
|
||||
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
||||
subinterval of $P$.
|
||||
Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
|
||||
|
||||
Let $c$ be a real number.
|
||||
Then $t(x) = s(x - c)$ is a step function on closed interval $[a + c, b + c]$
|
||||
with partition $Q = \{x_0 + c, x_1 + c, \ldots, x_n + c\}$.
|
||||
Furthermore, $t$ is constant on each open subinterval of $Q$.
|
||||
Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$.
|
||||
By construction, $t_k = s_k$.
|
||||
|
||||
By definition of the \nameref{sec:def-integral-step-function},
|
||||
\begin{align*}
|
||||
\int_{a+c}^{b+c} s(x - c) \mathop{dx}
|
||||
& = \int_{a+c}^{b+c} t(x) \mathop{dx} \\
|
||||
& = \sum_{k=1}^n t_k \cdot ((x_k + c) - (x_{k - 1} + c)) \\
|
||||
& = \sum_{k=1}^n t_k \cdot (x_k - x_{k - 1}) \\
|
||||
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k - 1}) \\
|
||||
& = \int_a^b s(x) \mathop{dx}.
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section{\partial{Expansion or Contraction of the Interval of Integration}}%
|
||||
\label{sec:step-expansion-contraction-interval-integration}
|
||||
|
||||
Let $s$ be a step function on closed interval $[a, b]$.
|
||||
Then
|
||||
$$\int_{ka}^{kb} s \left( \frac{x}{k} \right) \mathop{dx} =
|
||||
k \int_a^b s(x) \mathop{dx} \quad\text{for every } k > 0.$$
|
||||
|
||||
\todo{This also holds for negative values of $k$.}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $s$ be a step function on closed interval $[a, b]$.
|
||||
By definition of a step function, there exists a \nameref{sec:def-partition}
|
||||
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
||||
subinterval of $P$.
|
||||
Let $s_i$ denote the value of $s$ on the $i$th open subinterval of $P$.
|
||||
|
||||
Let $k > 0$ be a real number.
|
||||
Then $t(x) = s(x / k)$ is a step function on closed interval $[ka, kb]$
|
||||
with partition $Q = \{kx_0, kx_1, \ldots, kx_n\}$.
|
||||
Furthermore, $t$ is constant on each open subinterval of $Q$.
|
||||
Let $t_i$ denote the value of $t$ on the $i$th open subinterval of $Q$.
|
||||
By construction, $t_i = s_i$.
|
||||
|
||||
By definition of the \nameref{sec:def-integral-step-function},
|
||||
\begin{align*}
|
||||
\int_{ka}^{kb} s(x / k) \mathop{dx}
|
||||
& = \int_{ka}^{kb} t(x) \mathop{dx} \\
|
||||
& = \sum_{i=1}^n t_i \cdot (kx_i - kx_{i-1}) \\
|
||||
& = k \sum_{i=1}^n t_i \cdot (x_i - x_{i-1}) \\
|
||||
& = k \sum_{i=1}^n s_i \cdot(x_i - x_{i-1}) \\
|
||||
& = k \int_a^b s(x) \mathop{dx}.
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
19
preamble.tex
19
preamble.tex
|
|
@ -23,30 +23,27 @@
|
|||
% Environments
|
||||
% ========================================
|
||||
|
||||
\newcommand{\divider}{\vspace{10pt}\hrule\vspace{10pt}}
|
||||
\newcommand{\header}[2]{\title{#1}\author{#2}\date{}\maketitle}
|
||||
|
||||
\newenvironment{axiom}{%
|
||||
\paragraph{\normalfont\normalsize\textit{Axiom.}}}
|
||||
{\hfill$\square$}
|
||||
\newenvironment{definition}{%
|
||||
\paragraph{\normalfont\normalsize\textit{Definition.}}}
|
||||
{\hfill$\square$}
|
||||
\newcommand{\divider}{%
|
||||
\vspace{10pt}
|
||||
\hrule
|
||||
\vspace{10pt}}
|
||||
\newcommand{\header}[2]{%
|
||||
\title{#1}
|
||||
\author{#2}
|
||||
\date{}
|
||||
\maketitle}
|
||||
\newcommand{\note}[1]{%
|
||||
|
||||
\newcommand{\admonition}[2]{%
|
||||
\begin{center}
|
||||
\doublebox{
|
||||
\begin{minipage}{0.95\textwidth}
|
||||
\vspace{2pt}
|
||||
\hl{Note:} #1
|
||||
\hl{#1} #2
|
||||
\vspace{2pt}
|
||||
\end{minipage}}
|
||||
\end{center}}
|
||||
\newcommand{\note}[1]{\admonition{Note:}{#1}}
|
||||
\newcommand{\todo}[1]{\admonition{TODO:}{#1}}
|
||||
|
||||
% ========================================
|
||||
% Status
|
||||
|
|
|
|||
Loading…
Reference in New Issue