How to prove that if $f$ is integrable, then $\forall \epsilon >0, \ \exists$ partition $M\in [a,b]$ such that $U_f(M) - L_f(M)\lt\epsilon$?

Question

Here is my proof:

Since $f$ is integrable, $\overline{I}_a^b(f)$=$\underline{I}_a^b(f)$. However, it is also a fact that

$L_f(M) \leqslant \underline{I}_a^b(f) \leqslant \overline{I}_a^b(f) \leqslant U_f(M)$, meaning that there exists a $\epsilon>0$ such that $\overline{I}_a^b(f) \leqslant \underline{I}_a^b(f)+\epsilon$,

and therefore $0\leqslant \overline{I}_a^b(f)-\underline{I}_a^b(f)\leqslant \epsilon$. By the general inequality I pointed above, since the lower

integral is the supremum of lower sums $L_f(M)$ over all partitions and the upper integral is the

infimum of upper sums $U_f(M)$ over all partitions, $U_f(M)-L_f(M)<\epsilon, \ \ \forall \epsilon>0$

Could anyone correct my proof as I am struggling with the $\geqslant \gt$ signs?

Answer 1

Let's let $I$ be the integral of $f$, which exists and is finite by assumption. Because it is the sup of the lower sums, for any $\epsilon_1>0$ there are infinitely many partitions $M$ so that $I-L_f(M)<\epsilon_1$. Similarly, there are infinitely many partitions $N$ so that $U_f(N)-I<\epsilon_2$. If $\epsilon>0$, and we choose $M$ and $N$ with epsilons given by $\epsilon_1=\epsilon_2=\epsilon/2$, and let $P$ be the common refinement of $M$ and $N$, what can we say about $U_f(P)-L_f(P)$?

Answer 2

It is not clear your final argument:

since the lower integral is the supremum of lower sums $L_f(M)$ over all partitions and the upper integral is the infimum of upper sums $U_f(M)$ over all partitions, $U_f(M)−L_f(M)<ϵ$, $∀ϵ>0$

It seems that you are using that if $\alpha=\sup A=\inf B$, then there is $a\in A$ and $b\in B$ such that $b-a<\epsilon$. It is true, but not sufficient to conlude.

Now, since $\overline{I_a^b}(f)=\inf\{U_f(M)\}$, then $\overline{I_a^b}(f)+\frac{\epsilon}{2}$ is not a lower bound for $\inf\{U_f(M)\}$, so there is a partition $M_1$ such that $\overline{I_a^b}(f)\le U_f(M_1)<\overline{I_a^b}(f)+\frac{\epsilon}{2}$.

In the same way, there is a partition $M_2$ such that $\underline{I_a^b}-\frac{\epsilon}{2}<L_f(M_2)\le\underline{I_a^b}(f)$.

Let $M=M_1\cup M_2$. Thus

$$\underline{I_a^b}(f)-\frac{\epsilon}{2}<L_f(M_2)\le L_f(M)\le\underline{I_a^b}(f)\le\overline{I_a^b}(f)\le U_f(M)\le U_f(M_1)<\overline{I_a^b}(f)+\frac{\epsilon}{2}$$

Therefore $U_f(M)-L_f(M)<\overline{I_a^b}(f)+\frac{\epsilon}{2}-(\underline{I_a^b}(f)-\frac{\epsilon}{2})=\epsilon$ since $\overline{I_a^b}(f)=\underline{I_a^b}(f)$