I'm reading Bressoud's A radical approach to Lebesgue theory of integration and there's a section that I don't get, please read below:
Is there a mean value theorem for integrable functions ? I know there's one for integrals of continuous functions. If the continuity assumption is dropped, I don't know what to do…
Best Answer
Suppose $f$ is integrable (not necessarily continuous) and $\lim_{x \to a+} f(x)= L$. For any $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x) - L| < \epsilon$ when $0 < x < a + \delta$.
If $0 < h < \delta$, then
$$\left|\frac{1}{h}\int_a^{a+h} f(x) \, dx - L \right| = \left|\frac{1}{h}\int_a^{a+h} (f(x) - L) \, dx \right| \leqslant\frac{1}{h}\int_a^{a+h} |f(x) - L| \, dx < \epsilon $$
Thus, $\displaystyle \lim_{h \to 0+} \frac{1}{h}\int_a^{a+h} f(x) \, dx = L$.
A similar argument applies to the left-hand limit. The mean value theorem for integrals is not needed (nor does it apply to discontinuous functions.)