Indefinite Lebesgue Integrals on a Closed Interval

In the setting of Integrals over Subintervals of a Closed Interval: Notation, assume the axiom of dependent choice. For a function $g:[a,b]\to\mathbb{R}^{n}$ and $k\in[n]$, $g_{k}$ is its $k$-th component function.

(i) Restrictions to subintervals are integrable. Let $f\in\mathcal{L}^{1}([a,b])$, $g\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ and $a\le s\le t\le b$. Then $f|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]})$ and $g|_{[s,t]}\in\mathcal{L}^{1}(\lambda_{[s,t]};\mathbb{R}^{n})$, the product $f\mathbf{1}_{[s,t]}$ lies in $\mathcal{L}^{1}([a,b])$, and

$$\int_{[s,t]}f\,d\lambda=\int f\mathbf{1}_{[s,t]}\,d\lambda_{[a,b]},\qquad\Bigl(\int_{[s,t]}g\,d\lambda\Bigr)_{k}=\int_{[s,t]}g_{k}\,d\lambda\quad(k\in[n]).$$

For the rest of this statement, let $f\in\mathcal{L}^{1}([a,b])$ and $g\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$, let $\lVert g\rVert$ be the function $r\mapsto\lVert g(r)\rVert$ on $[a,b]$, and let $F:[a,b]\to\mathbb{R}$ and $G:[a,b]\to\mathbb{R}^{n}$ be the functions (Functions and Function-Valued Operators Given by Integrals over Subintervals (i) and Functions and Function-Valued Operators Given by Integrals over Subintervals (ii))

$$F(t)=\int_{[a,t]}f\,d\lambda,\qquad G(t)=\int_{[a,t]}g\,d\lambda\qquad(t\in[a,b]).$$

Here $|f|$ and $\lVert g\rVert$ lie in $\mathcal{L}^{1}([a,b])$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x) and Rules for Integrals of Vector- and Matrix-Valued Functions (viii)), so their integrals over $[s,t]$ are defined by (i).

(ii) Degenerate intervals. For every $s\in[a,b]$,

$$\int_{[s,s]}f\,d\lambda=0 .$$

(iii) Additivity. For all $r,s,t$ with $a\le r\le s\le t\le b$,

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

(iv) Monotonicity. Let $a\le s\le t\le b$, and let $h\in\mathcal{L}^{1}([a,b])$ with $f(r)\le h(r)$ for every $r\in[s,t]$. Then

$$\int_{[s,t]}f\,d\lambda\le\int_{[s,t]}h\,d\lambda,\qquad\Bigl|\int_{[s,t]}f\,d\lambda\Bigr|\le\int_{[s,t]}|f|\,d\lambda .$$

(v) Continuity. $F$ is uniformly continuous on $[a,b]$, and in particular continuous on $[a,b]$.

(vi) Lipschitz bound. If $K\in\mathbb{R}$ and $|f(r)|\le K$ for every $r\in[a,b]$, then

$$|F(t)-F(s)|\le K|t-s|\qquad(s,t\in[a,b]).$$

(vii) Norm bound for vector-valued integrands. For all $s,t$ with $a\le s\le t\le b$,

$$\Bigl\lVert\int_{[s,t]}g\,d\lambda\Bigr\rVert\le\int_{[s,t]}\lVert g\rVert\,d\lambda .$$

(viii) Vector-valued integrands: degenerate intervals and additivity. For every $s\in[a,b]$, $\int_{[s,s]}g\,d\lambda=0$, and for all $r,s,t$ with $a\le r\le s\le t\le b$,

$$\int_{[r,t]}g\,d\lambda=\int_{[r,s]}g\,d\lambda+\int_{[s,t]}g\,d\lambda .$$

(ix) Vector-valued integrands: continuity. For the Euclidean distance on $\mathbb{R}^{n}$, $G$ is uniformly continuous on $[a,b]$: for every $\varepsilon>0$ there is $\delta>0$ with

$$\lVert G(t)-G(s)\rVert<\varepsilon\qquad(s,t\in[a,b],\ |t-s|<\delta).$$

In particular, $G$ is continuous on $[a,b]$.

(x) Vector-valued integrands: Lipschitz bound. If $K\in\mathbb{R}$ and $\lVert g(r)\rVert\le K$ for every $r\in[a,b]$, then

$$\lVert G(t)-G(s)\rVert\le K|t-s|\qquad(s,t\in[a,b]).$$

(xi) Continuous integrands. Let $\varphi:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$. Then $\varphi\in\mathcal{L}^{1}([a,b])$, and for all $s,t$ with $a\le s\le t\le b$,

$$\int_{[s,t]}\varphi\,d\lambda=\int_{s}^{t}\varphi ,$$

the right-hand side being the oriented Riemann integral, defined by Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative (iv).