Riemann integrable function on each compact

Question

Let $f$ be a bounded function on $[a,b]$ which is Riemann integrable on each $[\alpha,\beta]\subset ]a,b[$

How to prove that f is Riemann integrable on [a,b]?

using sum of Darboux

Answer 1

The (usual) way of thinking is, if you only have information about the tiny compact intervals, try to express your current problem, in this case showing that a global integral exists, in terms of the integrals on the tiny compact intervals.

We'll show that $\lim_{n\to\infty}\overline{\int_{a+1/n}^{b-1/n}}f=\overline{\int_a^b}f$.

Since $f$ is bounded, the upper integral exists, that is, the upper integral is a real number.

Let $\varepsilon>0$. By the definition of upper integral, there's a $\delta>0$ such that if $$ P=\{t_0=a<t_1<\dots<t_n=b\} $$ is a partition of $[a,b]$ with $\operatorname{diam}(P)<\delta$, we have $$ U_P-\overline{\int_a^b}f < \varepsilon . $$

Let $n$ be such that $a<a+1/n<t_1$ and $t_{n-1}<b-1/n<b$. Next we apply the definition of the upper integral for a smaller compact interval. For $\varepsilon$ there's a $\delta'>0$ such that for any partition $P'$ of the interval $[a-1/n,b+1/n]$ we have $$ U_{P'}-\overline{\int_{a+1/n}^{b-1/n}}f < \varepsilon . $$

Take $\eta=\min\{\delta,\delta'\}$ and let $Q$ be the older partition $P$ with (a finite amount) of dummy points added so that $\operatorname{diam}(Q)<\eta$. Similarly, let $Q'=Q\cup\{a+1/n,b-1/n\}$. Note that $$ |U_Q-U_{Q'}|\leq M[(t_1-(a+1/n))+(a+1/n-a) + b-1/n-t_{n-1}+b-(b+1/n)] =M[t_1-a]+M[b-t_{n-1}]\leq 2\eta M , $$ where $M$ is a bound for $|f|$.

Finally, we estimate the difference $$ \Big|\overline{\int_a^b}f-\overline{\int_{a-1/n}^{b-1/n}}f\Big| \leq \Big|\overline{\int_a^b}f-U_Q\Big| + |U_Q-U_{Q'}| + \Big|U_{Q'}-\overline{\int_{a-1/n}^{b-1/n}}f\Big| \leq \varepsilon+2\eta M+\varepsilon . $$

What's left to do is do the same for the lower integral and put all the pieces together: $$ \overline{\int_a^b}f=\lim_{n\to\infty}\overline{\int_{a+1/n}^{b-1/n}}f =\lim_{n\to\infty}\underline{\int_{a+1/n}^{b-1/n}}f=\underline{\int_a^b}f . $$

I'll leave the trimming and decoration (adding details, I mean) of the above sketch to you.