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Let $ -\infty < A < B \leq +\infty$, and let $f:[A,B)\to \mathbb{R}$ be almost continuous everywhere (sucht that $f_{[A,b]} \in \mathcal{R}(A,b)$ for all $A<b<B$). Then $\mathcal{R} \int_A^B f(x)dx := \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx.$

The following is a condition when the improper Riemann integral is the same as the Lebesgue integral. How do I prove this?

If $f \geq 0$, then

$$\mathcal{L} \int_{[A,B)} fd\lambda = \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx \in \mathbb {\overline R}_{\ge 0}.$$

Maria M.
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1 Answers1

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Well, by definition if a function is riemann integrable then $$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$$

Notice that the function $g_n(x)=\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$ if $x\in \left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right)$ is increasing and converges to $f$ almost everywhere, also

$$\int_{[a,b]}g_{n}d\lambda=\frac{1}{n}\sum_{i=1}^n\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$$ And by the bounded convergence theorem we have that $\int_{[a,b]}fd\lambda=\lim\int_{[a.b]}g_nd\lambda=\int_a^bf(x)dx$

Now take $f:[A,B)\to\mathbb R$ and an increasing sequence $b_n\to B$ $$\int_{[A,B)}f\chi_{[A,b_n]}d\lambda=\int_{[A,b_n]}fd\lambda=\int_A^{b_n}f(x)dx$$ So by bounded convergence $\int_{[A,B)}fd\lambda=\lim\int_{[A,B)}f\chi_{[A,b_n]}d\lambda=\lim\int_A^{b_n}f(x)dx=\lim_{b\to B}\int_A^bf(x)dx$ Where the last equality holds because we are asuming that $\lim_{b\to B}\int_A^bf(x)dx$ conveges so for every sequence $b_n\to B$ $\int_A^{b_n}f(x)dx$ must have the same limit.

Matías Ures
  • 413