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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 $g \geq 0$ with $g \in \mathcal{L}([A,B))$ and $|f| \leq g$, then

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

Maria M.
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1 Answers1

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If $ f $ is almost-everywhere continuous then $ \mathcal{R} \int_A^b{f(x) dx} = \mathcal{L} \int_A^b{f(x) dx} $ for all $ A < b < B $. So the only thing you have to prove is $$ (*)\quad \lim_{b \rightarrow B} {\mathcal{L}\int_A^b{f(x) dx}} = \mathcal{L}\int_A^B{f(x) dx} $$

For this, note $ \mathcal{L}\int_A^b{f(x) dx} = \mathcal{L} \int_A^B{f(x) \chi_{[A,b)} dx} $. Now letting $ b_n $ be any sequence $ b_n \rightarrow B $, and $ f_n(x) = f(x)\chi_{[A,b_n)} $, we have $ |f_n(x)| < g(x) $ and $ f_n(x) \rightarrow f(x) $ (almost everywhere) on $ [A,B] $. The dominated convergence theorem immediately gives us the desired limit $ (*) $.

Jake Mirra
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