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. Let $f:[a,b]\to\mathbb{R}$ be continuous. By claims 2 and 3 of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, $\tilde f$ is $\mathcal{B}$-measurable and $\lambda$-integrable with $\int_{\mathbb{R}}\tilde f\,d\lambda$ equal to the Riemann integral of $f$. By claim 2 above, $f$ is $\mathcal{B}_{[a,b]}$-measurable and $\int_{[a,b]}f\,d\lambda_{[a,b]}=\int_{\mathbb{R}}\tilde f\,d\lambda$, proving the displayed equality; in particular $f$ is a random variable on the probability space of claim 1. By Extreme Value Theorem on a Compact Interval there is $M\ge0$ with $|f|\le M$ on $[a,b]$; $f^{2}$ is continuous by Sums and Products of Continuous Real-Valued Functions (in its one-dimensional instance), so by the same argument $f^{2}$ is measurable with $\int_{[a,b]}f^{2}\,d\lambda_{[a,b]}$ equal to its Riemann integral, which is at most $M^{2}(b-a)$ by claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals applied to the continuous function $f^{2}$ on $[a,b]$. By the scaling identity, the expectation of $f^{2}$ on the normalized space 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.