How to prove that $f$ is integrable if $\forall \epsilon, \ \exists$ partition $M\in [a,b]$ such that $U_f(M) - L_f(M)\lt\epsilon$?

Question

In order to let $f$ be integrable, its lower integral,$\underline{I}_a^b(f)$, has to equal upper integrable, $\overline{I}_a^b(f)$. I suppose that we can claim that because $\forall \epsilon, U_f(M) - L_f(M)\lt\epsilon$, $\ U_f(M) - L_f(M)=0$ would be the only solution since $\ U_f(M) - L_f(M)\ge 0$. So if $\ U_f(M) - L_f(M)=0$, $\underline{I}_a^b(f)$ has to be the greatest $L_f(M)$ and $\overline{I}_a^b(f)$ is the smallest $U_f(M)$. However I have a bit problem formalizing th proof and I am not sure which parts are wrong, so any thoughts would be greatly helpful.

Answer 1

The hypothesis does not allow you to conclude that $U_f(M) - L_f(M) < \epsilon$ for all $\epsilon>0$ and $M$ fixed. Rather for each $\epsilon > 0$ there is a partition $M_\epsilon$ for which $U_f(M_{\epsilon}) - L_f(M_{\epsilon}) < \epsilon.$

Nevertheless this implies that the difference between upper and lower integrals can be squeezed between $0$ and any $\epsilon > 0$, hence equal to $0$. The lower integral is the supremum of lower sums over all partitions and the upper integral is the infimum of upper sums over all partitions. Hence,

$$L_f(M_{\epsilon}) \leqslant \underline{I}_a^b(f) \leqslant \overline{I}_a^b(f) \leqslant U_f(M_{\epsilon}) \\ \implies 0 \leqslant \overline{I}_a^b(f) - \underline{I}_a^b(f) \leqslant U_f(M_{\epsilon}) - L_f(M_{\epsilon}) \leqslant \epsilon \\ \implies \overline{I}_a^b(f) = \underline{I}_a^b(f) $$