Proof of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval

Throughout, $\mathcal{B}$, $\lambda$, $\mathcal{B}_{[a,b]}$, $\lambda_{[a,b]}$, and zero extensions are as in the statement. We use that the Borel $\sigma$-algebra contains every open subset of $\mathbb{R}$ and, by closure of a $\sigma$-algebra under complements, every closed subset; in particular $[a,b]\in\mathcal{B}$. We also record for repeated use the monotonicity of a measure: if $E\subseteq F$ are members of a $\sigma$-algebra on which $\mu$ is a measure, then $\mu(E)\le\mu(F)$, since $\mu(F)=\mu(E)+\mu(F\setminus E)$ by additivity, with nonnegative terms.

Claim 1. Every member of $\mathcal{B}_{[a,b]}$ has the form $S\cap[a,b]$ with $S\in\mathcal{B}$ and hence lies in $\mathcal{B}$, being an intersection of two members of $\mathcal{B}$. The collection $\mathcal{B}_{[a,b]}$ is a $\sigma$-algebra on $[a,b]$: it contains $[a,b]=\mathbb{R}\cap[a,b]$; if $E=S\cap[a,b]$ then $[a,b]\setminus E=(\mathbb{R}\setminus S)\cap[a,b]\in\mathcal{B}_{[a,b]}$; and if $E_n=S_n\cap[a,b]$ for $n\in\mathbb{N}$ then $\bigcup_nE_n=\bigl(\bigcup_nS_n\bigr)\cap[a,b]\in\mathcal{B}_{[a,b]}$. The set function $\lambda_{[a,b]}$ is a measure: $\lambda_{[a,b]}(\emptyset)=\lambda(\emptyset)=0$, and for pairwise disjoint $E_n\in\mathcal{B}_{[a,b]}\subseteq\mathcal{B}$, countable additivity of $\lambda$ gives $\lambda_{[a,b]}\bigl(\bigcup_nE_n\bigr)=\sum_n\lambda_{[a,b]}(E_n)$. By claim 4 of Existence of Lebesgue Measure on the Real Line, $\lambda([a,b])=b-a$. Finally, multiplying a measure by the positive constant $(b-a)^{-1}$ preserves $\mu(\emptyset)=0$ and countable additivity, and gives total mass $(b-a)^{-1}(b-a)=1$, so $([a,b],\mathcal{B}_{[a,b]},(b-a)^{-1}\lambda_{[a,b]})$ is a probability space.

We record a scaling identity used below: for every constant $c>0$, every $\mathcal{B}_{[a,b]}$-measurable $h:[a,b]\to[0,\infty]$ satisfies $\int h\,d(c\,\lambda_{[a,b]})=c\int h\,d\lambda_{[a,b]}$. Indeed, by Simple Function and Its Integral the integral of a simple function $s=\sum_ic_i\mathbf{1}_{A_i}$ with respect to $c\,\lambda_{[a,b]}$ is $\sum_ic_i\,c\,\lambda_{[a,b]}(A_i)$, which is $c$ times its integral with respect to $\lambda_{[a,b]}$; the class of simple functions below $h$ is the same for both measures, and the supremum defining the integral scales by $c$.

Claim 2. Let $f:[a,b]\to[0,\infty]$ and let $A$ be a Borel subset of $[0,\infty]$ in the sense of Lebesgue Integral of a Nonnegative Measurable Function. If $0\notin A$ then $\tilde f^{-1}(A)=f^{-1}(A)$, and if $0\in A$ then $\tilde f^{-1}(A)=f^{-1}(A)\cup(\mathbb{R}\setminus[a,b])$. If $f$ is $\mathcal{B}_{[a,b]}$-measurable then $f^{-1}(A)\in\mathcal{B}_{[a,b]}\subseteq\mathcal{B}$ and $\mathbb{R}\setminus[a,b]\in\mathcal{B}$, so $\tilde f$ is $\mathcal{B}$-measurable. Conversely, if $\tilde f$ is $\mathcal{B}$-measurable then $f^{-1}(A)=\tilde f^{-1}(A)\cap[a,b]\in\mathcal{B}_{[a,b]}$.

For the integrals, we set up a correspondence of simple functions. If $s$ is a simple function on $([a,b],\mathcal{B}_{[a,b]})$ with $s\le f$, its zero extension $\tilde s$ is simple on $(\mathbb{R},\mathcal{B})$ with $\tilde s\le\tilde f$, and by Simple Function and Its Integral the two integrals agree, the added value $0$ on $\mathbb{R}\setminus[a,b]$ contributing $0$ by the convention $0\cdot\infty=0$ of Measure, Measure Space, and Probability Measure. Conversely, if $s'$ is simple on $(\mathbb{R},\mathcal{B})$ with $s'\le\tilde f$, then $s'=0$ on $\mathbb{R}\setminus[a,b]$ because $\tilde f=0$ there and $s'\ge0$; hence $s'$ is the zero extension of its restriction to $[a,b]$, which is simple with $s'\restriction_{[a,b]}\le f$ and has the same integral. The two suprema in Lebesgue Integral of a Nonnegative Measurable Function therefore coincide, which is the asserted equality. For real-valued $f$, apply the above to the positive and negative parts, noting $\widetilde{(f^{+})}=(\tilde f)^{+}$ and $\widetilde{(f^{-})}=(\tilde f)^{-}$, and use Integrable Function and the Lebesgue Integral.

Claim 3. Regard $[a,b]$ as a subset of the real line $(\mathbb{R},d_{\mathbb{R}})$, so that $d_{\mathbb{R}}(s,t)=|s-t|$ for all $s,t\in[a,b]$, and let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$. The existence argument below adapts a previously published TheoremBase argument to the present continuity hypothesis; see the attached citation.

Uniform continuity. By Closed Interval $[a,b]$ is Compact in $\mathbb{R}$ the interval $[a,b]$ is a compact subset of the real line. Hence Heine-Cantor Theorem: Continuity on a Compact Subset Implies Uniform Continuity, applied with both metric spaces equal to the real line and with $K=[a,b]$, shows that $f$ is uniformly continuous on $[a,b]$: for every $\varepsilon>0$ there is $\delta>0$ such that all $s,t\in[a,b]$ with $d_{\mathbb{R}}(s,t)<\delta$ satisfy $|f(s)-f(t)|<\varepsilon$. Since $d_{\mathbb{R}}(s,t)=|s-t|$, this says precisely that $f$ is uniformly continuous on the subset $[a,b]$ of $\mathbb{R}$, and we use it in this form below.

Boundedness. Apply uniform continuity with $\varepsilon=1$ to obtain $\delta_{1}>0$. By claim 3 of The Archimedean Property of the Real Numbers there is a natural number $k$ with $(b-a)/k<\delta_{1}$, and $k\le2^{k}$ by an easy induction, so $n_{1}=k$ satisfies $(b-a)/2^{n_{1}}<\delta_{1}$. Put $x_{i}=a+i(b-a)/2^{n_{1}}$ for $i=0,\dots,2^{n_{1}}$. Every $t\in[a,b]$ lies in $[x_{i-1},x_{i}]$ for the least $i\ge1$ with $t\le x_{i}$ — such an $i$ exists since $t\le b=x_{2^{n_{1}}}$, and for $t=a$ it is $i=1$ — and then $|t-x_{i-1}|\le(b-a)/2^{n_{1}}<\delta_{1}$, so $|f(t)|\le|f(t)-f(x_{i-1})|+|f(x_{i-1})|<1+\max\{|f(x_{j})|:0\le j\le2^{n_{1}}\}$. Write $M$ for the right-hand side, a real number: then $|f(t)|\le M$ for every $t\in[a,b]$.

Riemann integrability. For each natural number $n$ let $P_{n}$ be the partition of $[a,b]$ with division points $x^{(n)}_{i}=a+i(b-a)/2^{n}$, $i=0,\dots,2^{n}$; its mesh, the largest of its subinterval lengths, is $(b-a)/2^{n}$. Let $\mathcal{T}_{n}$ be the tagged partition on $P_{n}$ assigning to each subinterval its left endpoint as tag, and let $R_{n}=\sum_{i=1}^{2^{n}}f(x^{(n)}_{i-1})\,(x^{(n)}_{i}-x^{(n)}_{i-1})$ be the corresponding Riemann sum of $f$. For a tagged partition $\mathcal{W}$ of $[a,b]$ relative to a partition of $[a,b]$, write $R(f,\mathcal{W})$ for its Riemann sum of $f$, and call the mesh of the underlying partition the mesh of $\mathcal{W}$; thus $R_{n}=R(f,\mathcal{T}_{n})$.

We record the key estimate. Let $\varepsilon>0$ with $\delta>0$ as in the uniform continuity statement, and let $\mathcal{U}$ and $\mathcal{V}$ be tagged partitions of $[a,b]$ whose meshes are both less than $\delta/2$. Then $|R(f,\mathcal{U})-R(f,\mathcal{V})|\le\varepsilon(b-a)$. To see this, let $Q$ be the common refinement of the two underlying partitions — the partition of $[a,b]$ whose set of division points is the union of the two sets of division points, a finite subset of $[a,b]$ containing $a$ and $b$, hence again a partition — and list its subintervals as $[y_{j-1},y_{j}]$, $j=1,\dots,J$. Each subinterval of $\mathcal{U}$'s partition is the union of consecutive subintervals of $Q$, whose lengths sum to its own length; grouping the terms accordingly shows that

where $\tau_{j}$ is the tag of the subinterval of $\mathcal{U}$ containing $[y_{j-1},y_{j}]$, and likewise $R(f,\mathcal{V})=\sum_{j=1}^{J}f(\tau'_{j})\,(y_{j}-y_{j-1})$ with $\tau'_{j}$ the tag of the containing subinterval of $\mathcal{V}$. For each $j$, picking any $y\in[y_{j-1},y_{j}]$, the numbers $\tau_{j}$ and $y$ lie in one subinterval of $\mathcal{U}$'s partition and $\tau'_{j}$ and $y$ in one of $\mathcal{V}$'s, so $|\tau_{j}-\tau'_{j}|\le|\tau_{j}-y|+|y-\tau'_{j}|<\delta/2+\delta/2=\delta$, whence $|f(\tau_{j})-f(\tau'_{j})|<\varepsilon$. Summing, $|R(f,\mathcal{U})-R(f,\mathcal{V})|\le\varepsilon\sum_{j}(y_{j}-y_{j-1})=\varepsilon(b-a)$.

The sequence $(R_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence: given $\varepsilon'>0$, apply the key estimate with $\varepsilon=\varepsilon'/(b-a+1)$, whose associated $\delta$ yields, for all $m,n$ with meshes of $P_{m}$ and $P_{n}$ less than $\delta/2$, the bound $|R_{n}-R_{m}|\le\varepsilon'(b-a)/(b-a+1)<\varepsilon'$; the mesh condition holds for all sufficiently large $m,n$ as in the boundedness step. By Every Cauchy Sequence of Real Numbers Converges there is $I\in\mathbb{R}$ such that $(R_{n})$ converges to $I$.

Now let $\eta>0$. Apply the key estimate with $\varepsilon=\eta/(b-a+1)$ and its $\delta$. If $\mathcal{U}$ is any tagged partition of $[a,b]$ with mesh less than $\delta/2$, then for every $n$ large enough that the mesh of $P_{n}$ is less than $\delta/2$ we get $|R(f,\mathcal{U})-R_{n}|\le\eta(b-a)/(b-a+1)$. Let $\theta>0$; since $(R_{n})$ converges to $I$, there is such an $n$ with additionally $|R_{n}-I|<\theta$, and then $|R(f,\mathcal{U})-I|\le|R(f,\mathcal{U})-R_{n}|+|R_{n}-I|<\eta(b-a)/(b-a+1)+\theta$. As $\theta>0$ was arbitrary, $|R(f,\mathcal{U})-I|\le\eta(b-a)/(b-a+1)$ — otherwise taking $\theta$ equal to the positive difference of the two sides gives a contradiction — and hence $|R(f,\mathcal{U})-I|<\eta$. By Riemann Integrability on a Closed Interval, $f$ is Riemann integrable on $[a,b]$ and $\int_{a}^{b}f(t)\,dt=I$.

Measurability, integrability, and agreement of the integrals. By claim 4 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions, $f$ is measurable with respect to the trace Borel $\sigma$-algebra on $[a,b]$, which is $\mathcal{B}_{[a,b]}$, and the Borel $\sigma$-algebra on the real line; in particular $f$ is a random variable on the probability space of claim 1. Its positive and negative parts $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$ are measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable, applied to the sequentially continuous maps $x\mapsto\max(x,0)$ and $x\mapsto\max(-x,0)$, and satisfy $0\le f^{\pm}\le M$. The constant function $M\mathbf{1}_{[a,b]}$ has integral $M\lambda_{[a,b]}([a,b])=M(b-a)$ by The Integral of an Indicator Function is the Measure of the Set and linearity, Linearity and Monotonicity of the Lebesgue Integral, and monotonicity, claim 1 there, gives $\int_{[a,b]}f^{\pm}\,d\lambda_{[a,b]}\le M(b-a)<\infty$; hence $f$ is integrable with respect to $\lambda_{[a,b]}$.

For each $n$ define $g_{n}=\sum_{i=1}^{2^{n}}f(x^{(n)}_{i-1})\,\mathbf{1}_{A^{(n)}_{i}}$, where $A^{(n)}_{i}=[x^{(n)}_{i-1},x^{(n)}_{i})$ for $i<2^{n}$ and $A^{(n)}_{2^{n}}=[x^{(n)}_{2^{n}-1},b]$. These sets are pairwise disjoint with union $[a,b]$, and each lies in $\mathcal{B}_{[a,b]}$: it is the intersection with $[a,b]$ of a closed interval, or of a closed interval with a singleton removed, and closed sets lie in $\mathcal{B}$ as recorded in the preamble. For every $x\in\mathbb{R}$ and $\eta>0$, monotonicity and claim 4 of Existence of Lebesgue Measure on the Real Line give $\lambda(\{x\})\le\lambda([x,x+\eta])=\eta$, so $\lambda(\{x\})=0$; additivity then gives $\lambda_{[a,b]}(A^{(n)}_{i})=x^{(n)}_{i}-x^{(n)}_{i-1}$ in all cases. By The Integral of an Indicator Function is the Measure of the Set and linearity, $g_{n}$ is integrable with

Let $\varepsilon>0$ with associated $\delta$ as in the uniform continuity statement, and let $n$ satisfy $(b-a)/2^{n}<\delta$. For $t\in A^{(n)}_{i}$ we have $|t-x^{(n)}_{i-1}|<\delta$ and hence $|f(t)-g_{n}(t)|=|f(t)-f(x^{(n)}_{i-1})|<\varepsilon$; thus $g_{n}-\varepsilon\mathbf{1}_{[a,b]}\le f\le g_{n}+\varepsilon\mathbf{1}_{[a,b]}$ pointwise. Monotonicity and linearity of the integral for integrable functions, Linearity and Monotonicity of the Lebesgue Integral, therefore give

for every such $n$. Let $\theta>0$; since $(R_{n})$ converges to $I$, there is such an $n$ with additionally $|R_{n}-I|<\theta$, and then $\bigl|\int_{[a,b]}f\,d\lambda_{[a,b]}-I\bigr|\le\varepsilon(b-a)+\theta$. As $\theta>0$ was arbitrary, $\bigl|\int_{[a,b]}f\,d\lambda_{[a,b]}-I\bigr|\le\varepsilon(b-a)$, by the contradiction argument used above; and as this holds for every $\varepsilon>0$, the left-hand side is smaller than every positive real number — given $\theta'>0$ take $\varepsilon=\theta'/(b-a+1)$ — so it is $0$, whence $\int_{[a,b]}f\,d\lambda_{[a,b]}=I=\int_{a}^{b}f(t)\,dt$, the Riemann integral. This proves the displayed equality of claim 3.

Square integrability. The map $f^{2}$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable, applied to the sequentially continuous map $x\mapsto x^{2}$, and $0\le f^{2}\le M^{2}$ pointwise, so as above $\int_{[a,b]}f^{2}\,d\lambda_{[a,b]}\le M^{2}(b-a)<\infty$. By the scaling identity, the expectation of $f^{2}$ on the normalized space of claim 1 is $(b-a)^{-1}\int_{[a,b]}f^{2}\,d\lambda_{[a,b]}<\infty$, so $f$ is square-integrable there.

Claim 4. Work on the probability space $([a,b],\mathcal{B}_{[a,b]},\mathbb{Q})$ of claim 1, $\mathbb{Q}:=(b-a)^{-1}\lambda_{[a,b]}$, and write $\mathbb{E}_{\mathbb{Q}}$ for its expectation. By the scaling identity, $\mathbb{E}_{\mathbb{Q}}[f^{2}]=(b-a)^{-1}\int_{[a,b]}f^{2}\,d\lambda_{[a,b]}<\infty$ and likewise for $g$, so $f$ and $g$ are square-integrable random variables on this space. By Square-Integrable Random Variables and the Mean-Square Inner Product the product $fg$ is $\mathbb{Q}$-integrable, hence $\lambda_{[a,b]}$-integrable by the scaling identity applied to $(fg)^{\pm}$, and claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm gives $\bigl|\mathbb{E}_{\mathbb{Q}}[fg]\bigr|^{2}\le\mathbb{E}_{\mathbb{Q}}[f^{2}]\,\mathbb{E}_{\mathbb{Q}}[g^{2}]$. Multiplying both sides by $(b-a)^{2}$ and using the scaling identity again yields the stated inequality. For the final assertion take $g=1$, which is measurable with $\int_{[a,b]}g^{2}\,d\lambda_{[a,b]}=b-a$, and replace $f$ by $|f|$, which satisfies $|f|^{2}=f^{2}$ and is measurable: for a Borel set $A\subseteq\mathbb{R}$,

and $-B:=\{-x:x\in B\}$ is Borel for every Borel $B$, because the collection of sets $B$ with $-B\in\mathcal{B}$ is a $\sigma$-algebra containing all open sets (the reflection of an open set is open).

Claim 5. On the probability space of claim 1, $f\ge0$ is a random variable with $\mathbb{E}_{\mathbb{Q}}[f]=(b-a)^{-1}\cdot0=0$ by the scaling identity. For every $n\ge1$, Markov's inequality (Markov's and Chebyshev's Inequalities) gives $\mathbb{Q}(f\ge1/n)\le n\,\mathbb{E}_{\mathbb{Q}}[f]=0$, so $\lambda_{[a,b]}(\{f\ge1/n\})=0$. Since $\{f>0\}=\bigcup_{n\ge1}\{f\ge1/n\}$, countable subadditivity gives $\lambda_{[a,b]}(\{f>0\})=0$; subadditivity follows from countable additivity by replacing $A_n:=\{f\ge1/n\}$ with the disjoint sets $B_n:=A_n\setminus\bigcup_{i<n}A_i$ and using $\lambda_{[a,b]}(B_n)\le\lambda_{[a,b]}(A_n)$, which is the monotonicity recorded in the preamble.

Claim 6. Write $N:=[a,b]\setminus D$, so $\lambda_{[a,b]}(N)=0$. For each $n$ the function $h_n:=f_n\mathbf{1}_D$ is $\mathcal{B}_{[a,b]}$-measurable: for a Borel set $A\subseteq[0,\infty]$, its preimage is $f_n^{-1}(A)\cap D$ if $0\notin A$ and $\bigl(f_n^{-1}(A)\cap D\bigr)\cup N$ if $0\in A$, and both belong to $\mathcal{B}_{[a,b]}$. For every $t\in[a,b]$ we have $h_n(t)\to f(t)$: on $D$ this is the hypothesis, and on $N$ both sides are $0$. Hence $f=\liminf_nh_n$ pointwise (nonnegative real-valued functions are in particular $[0,\infty]$-valued), so $f$ is measurable by Fatou's Lemma.

For the integral identity, let $g:[a,b]\to[0,\infty]$ be measurable. As in the previous paragraph, $g\mathbf{1}_D$ and $g\mathbf{1}_N$ are measurable, and $g=g\mathbf{1}_D+g\mathbf{1}_N$ pointwise, so by linearity of the integral of nonnegative measurable functions (Linearity and Monotonicity of the Lebesgue Integral) it suffices to show $\int_{[a,b]}g\mathbf{1}_N\,d\lambda_{[a,b]}=0$. Let $s=\sum_ic_i\mathbf{1}_{A_i}$ be any simple function with $0\le s\le g\mathbf{1}_N$, written with pairwise disjoint $A_i$ and, discarding zero terms, with every $c_i>0$. For $t\notin N$ we have $g\mathbf{1}_N(t)=0$, so $s(t)=0$ and therefore $A_i\subseteq N$ for every $i$; the monotonicity recorded in the preamble gives $\lambda_{[a,b]}(A_i)=0$, so the integral of $s$ is $0$. Taking the supremum over such $s$ in Lebesgue Integral of a Nonnegative Measurable Function yields $\int_{[a,b]}g\mathbf{1}_N\,d\lambda_{[a,b]}=0$, as required.