Proof of Indefinite Lebesgue Integrals on a Closed Interval

Throughout, $([a,b],\mathcal{B}_{[a,b]},\lambda_{[a,b]})$ is a measure space, by the axiom of dependent choice (Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness (iv) and Integrals of Functions over a Measurable Subset (i)); and for $a\le s\le t\le b$ the intervals $[s,t]$ and $(s,t]$ lie in $\mathcal{B}_{[a,b]}$, with

$$\lambda_{[a,b]}([s,t])=\lambda_{[a,b]}((s,t])=t-s$$

(Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness (i)).

Step 1: (i). Let $f$, $g$, $s$, $t$ be as there. By Integrals over Measurable Subsets (vi), $f$ is integrable on $[s,t]$ with respect to $\lambda_{[a,b]}$; as $\lambda_{[s,t]}$ is the restriction of $\lambda_{[a,b]}$ to $[s,t]$ (Integrals over Subintervals of a Closed Interval: Notation (iii)), this means $f|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]})$ (Integrals of Functions over a Measurable Subset (iii)), and the integral of $f$ over $[s,t]$ with respect to $\lambda_{[a,b]}$ (Integrals of Functions over a Measurable Subset (iv)) is $\int f|_{[s,t]}\,d\lambda_{[s,t]}=\int_{[s,t]}f\,d\lambda$. Now Integrals over Measurable Subsets (ii) gives $f\mathbf{1}_{[s,t]}\in\mathcal{L}^{1}([a,b])$ and $\int_{[s,t]}f\,d\lambda=\int f\mathbf{1}_{[s,t]}\,d\lambda_{[a,b]}$.

For $\psi\in\mathcal{L}^{1}([a,b])$ and any $B\in\mathcal{B}_{[a,b]}$ we write $\int_{B}\psi\,d\lambda$ for the integral of $\psi$ over $B$ with respect to $\lambda_{[a,b]}$; we have just seen that for $B=[s,t]$ this is the integral of Integrals over Subintervals of a Closed Interval: Notation (iii), so the rules of Integrals over Measurable Subsets apply to the integrals of the setting.

The restriction $g|_{[s,t]}$ is measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (viii)), and its components $g_{k}|_{[s,t]}$ lie in $\mathcal{L}^{1}(\lambda_{[s,t]})$ by the scalar case, since $g_{k}\in\mathcal{L}^{1}([a,b])$ (Integrable Vector- and Matrix-Valued Functions and Their Integrals (i)). Hence $g|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]};\mathbb{R}^{n})$, and by Integrable Vector- and Matrix-Valued Functions and Their Integrals (ii) the $k$-th component of $\int_{[s,t]}g\,d\lambda=\int g|_{[s,t]}\,d\lambda_{[s,t]}$ is

$$\int g_{k}|_{[s,t]}\,d\lambda_{[s,t]}=\int_{[s,t]}g_{k}\,d\lambda .$$

Step 2: (ii) and (iii). Since $\lambda_{[a,b]}([s,s])=0$, Integrals over Measurable Subsets (ix) gives $\int_{[s,s]}f\,d\lambda=0$. For $a\le r\le s\le t\le b$, $[r,t]$ is the disjoint union of $[r,s]$ and $(s,t]$, and $[s,t]$ is the disjoint union of $[s,s]$ and $(s,t]$. Additivity over disjoint sets (Integrals over Measurable Subsets (vii)) and (ii) give

$$\int_{[r,t]}f\,d\lambda=\int_{[r,s]}f\,d\lambda+\int_{(s,t]}f\,d\lambda,\qquad\int_{[s,t]}f\,d\lambda=\int_{[s,s]}f\,d\lambda+\int_{(s,t]}f\,d\lambda=\int_{(s,t]}f\,d\lambda,$$

and together these are (iii).

Step 3: (iv). In the measure space $([s,t],\mathcal{B}_{[s,t]},\lambda_{[s,t]})$ we have $f|_{[s,t]},h|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]})$ by (i), $f|_{[s,t]}\le h|_{[s,t]}$ pointwise, and $\lvert f|_{[s,t]}\rvert=\lvert f\rvert|_{[s,t]}$. Monotonicity and the absolute-value bound in this space (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix) and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x)) are therefore the two inequalities of (iv).

Step 4: (xi). By Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii), $\varphi\in\mathcal{L}^{1}([a,b])$. Let $a\le s\le t\le b$. If $s=t$, both sides vanish, by (ii) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii). Let $s<t$. By Integrals of Functions over a Measurable Subset (i), applied twice,

$$\mathcal{B}_{[s,t]}=\{B\in\mathcal{B}_{[a,b]}:B\subseteq[s,t]\}=\{B\in\mathcal{B}(\mathbb{R}):B\subseteq[s,t]\},\qquad\lambda_{[s,t]}(B)=\lambda_{[a,b]}(B)=\lambda(B)\quad(B\in\mathcal{B}_{[s,t]}),$$

so $([s,t],\mathcal{B}_{[s,t]},\lambda_{[s,t]})$ is also the restriction of $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ to $[s,t]$, the measure space of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval with $s,t$ in place of $a,b$. The restriction $\varphi|_{[s,t]}$ is continuous on $[s,t]$ (The Real Line: Standing Notation and Background for Calculus on Intervals (vi)), so Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii), applied with $s,t$ in place of $a,b$, and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi) give

$$\int_{[s,t]}\varphi\,d\lambda=\int\varphi|_{[s,t]}\,d\lambda_{[s,t]}=\int_{s}^{t}\varphi|_{[s,t]}=\int_{s}^{t}\varphi .$$

Step 5: (v) and (vi). For $a\le s\le t\le b$, (iii) with $r=a$ gives

$$F(t)-F(s)=\int_{[s,t]}f\,d\lambda .\tag{$*$}$$

Let $\varepsilon>0$, and let $\delta>0$ be given for $f$ and $\varepsilon$ by Absolute Continuity of the Integral (ii) in the measure space $([a,b],\mathcal{B}_{[a,b]},\lambda_{[a,b]})$. Let $y,z\in[a,b]$ with $|y-z|<\delta$, and let $s$ and $t$ be the smaller and the larger of them. Then $\lambda_{[a,b]}([s,t])=t-s<\delta$, so by $(*)$

$$|F(y)-F(z)|=\Bigl|\int_{[s,t]}f\,d\lambda\Bigr|<\varepsilon .$$

Thus $F$ is uniformly continuous on $[a,b]$, by the criterion of The Real Line: Standing Notation and Background for Calculus on Intervals (vi), and hence continuous on $[a,b]$ (A Uniformly Continuous Map Between Metric Spaces Is Continuous (i)).

Now let $|f(r)|\le K$ for every $r\in[a,b]$, and let $a\le s\le t\le b$. The constant function $K$ on $[a,b]$ is continuous, so by (xi) it lies in $\mathcal{L}^{1}([a,b])$ and $\int_{[s,t]}K\,d\lambda=\int_{s}^{t}K=K(t-s)$ (Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative (iii)). By $(*)$ and (iv), applied to $f$ and then to $|f|\le K$,

$$|F(t)-F(s)|\le\int_{[s,t]}|f|\,d\lambda\le\int_{[s,t]}K\,d\lambda=K(t-s)=K|t-s| .$$

Exchanging the roles of $s$ and $t$ gives the case $t<s$.

Step 6: the vector-valued clauses. Let $a\le s\le t\le b$. By (i), $g|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]};\mathbb{R}^{n})$, and $\lVert g|_{[s,t]}\rVert=\lVert g\rVert|_{[s,t]}$. The norm bound in $([s,t],\mathcal{B}_{[s,t]},\lambda_{[s,t]})$ (Rules for Integrals of Vector- and Matrix-Valued Functions (viii)) gives (vii):

$$\Bigl\lVert\int_{[s,t]}g\,d\lambda\Bigr\rVert=\Bigl\lVert\int g|_{[s,t]}\,d\lambda_{[s,t]}\Bigr\rVert\le\int\lVert g|_{[s,t]}\rVert\,d\lambda_{[s,t]}=\int_{[s,t]}\lVert g\rVert\,d\lambda .$$

(viii) is (ii) and (iii) applied to each component $g_{k}$, since by (i) the integrals of $g$ are formed componentwise, as are sums in $\mathbb{R}^{n}$.

Let $H:[a,b]\to\mathbb{R}$ be the function with $H(t)=\int_{[a,t]}\lVert g\rVert\,d\lambda$ (Functions and Function-Valued Operators Given by Integrals over Subintervals (i)). For $a\le s\le t\le b$, (viii) with $r=a$, (vii), and (iii) for $\lVert g\rVert$ give

$$0\le\lVert G(t)-G(s)\rVert=\Bigl\lVert\int_{[s,t]}g\,d\lambda\Bigr\rVert\le\int_{[s,t]}\lVert g\rVert\,d\lambda=H(t)-H(s).$$

In particular $H(t)-H(s)\ge0$, so $H$ is nondecreasing and $H(t)-H(s)=|H(t)-H(s)|$. As both sides are symmetric in $s$ and $t$, $\lVert G(y)-G(z)\rVert\le|H(y)-H(z)|$ for all $y,z\in[a,b]$, and $\lVert G(y)-G(z)\rVert$ is the Euclidean distance of $G(y)$ and $G(z)$ (The Euclidean Distance on $\mathbb{R}^n$ (i)). By (v) for $\lVert g\rVert$ and the criterion of The Real Line: Standing Notation and Background for Calculus on Intervals (vi), given $\varepsilon>0$ there is $\delta>0$ with $|H(y)-H(z)|<\varepsilon$ whenever $y,z\in[a,b]$ and $|y-z|<\delta$; the same $\delta$ serves for $G$. So $G$ is uniformly continuous on $[a,b]$ (Uniformly Continuous Map Between Metric Spaces (i)), and hence continuous on $[a,b]$ (A Uniformly Continuous Map Between Metric Spaces Is Continuous (i)), which is (ix). If $\lVert g(r)\rVert\le K$ for every $r\in[a,b]$, then (vi) for $\lVert g\rVert$ gives $|H(y)-H(z)|\le K|y-z|$, which proves (x).