# Show $f$ if integrable on $[a,b]$ if and only if $\epsilon > 0$, there exists a partition $P_{\epsilon}$ of [a,b]

Let $f$ be bounded on a nondegenerate interval $[a,b]$. Prove that $f$ if integrable on $[a,b]$ if and only if $\epsilon > 0$, there exists a partition $P_{\epsilon}$ of [a,b] such that P is a refinement of the partition $P_{\epsilon}$ implies $|U(f,P) - L(f,P)| < \epsilon$.

attempt $\rightarrow$: Suppose $f$ if integrable on $[a,b]$, then by definition there is a partition $P_{\epsilon}$ of [a,b], such that $|U(f,P_{\epsilon}) - L(f,P_{\epsilon})| < \epsilon$. Then let P be any partition such that P is a refinement of $P_{\epsilon}$, so P is finer than $P_{\epsilon}$ . Thus, $|U(f,P) - L(f,P)| < \epsilon$.

converse attempt: Suppose $\epsilon > 0$, there exists a partition $P_{\epsilon}$ of [a,b] such that P is a refinement of the partition $P_{\epsilon}$ implies $|U(f,P) - L(f,P)| < \epsilon$, then by definition $f$ is integrable, thus f is integrable.

Can anyone verify my proof? Any feedback/help would be helpful. For the converse I am not sure, I think it is trivial. Thank you so much.

• I'm confused. If you recast your proof as the general $A \iff B$ then it seems when you try to prove $A \implies B$, you assume $B$, effectively showing that $B \implies B$. Does that make sense? I think you need to assume $f$ is integrable under a different definition of integrable, and then show that definition implies the Partition-$\epsilon$ definition. – graydad Dec 4 '14 at 17:13
• – Martin Sleziak Oct 25 '15 at 7:40