Let $ f: [a, b] \to \mathbb{R}$ with the property that for any real number $E>0$ the set $A_E := \{x \in [a,b]: |f(x)| > E\}$ is finite or empty. Prove that $f$ is integrable.
My idea: I know that if the moduli of a Riemann sum is smaller than $E$, then from the epsilon-criterium it converges to $0$. Please help me solve it.
I assume you want to prove this for Riemann integrals without invoking measure theory.
For any $\epsilon > 0$ take $E = \frac{\epsilon}{4(b-a)}$.
There are only a finite number of points $x_1, \ldots, x_m \in A_E$. Cover these points with non-overlapping closed intervals $\bar{I}_j \subset [a,b]$ with lengths $l(\bar{I}_j) < \frac{\epsilon}{4m|f(x_j)|}$. Construct a partition $P$ of $[a,b]$ using the intervals $\bar{I}_1, \ldots ,\bar{I}_m$ and non-overlapping closed intervals $\bar{J}_1,\ldots, \bar{J}_n$ where $\bar{J}_j \cap A_E = \emptyset$.
The difference between upper and lower Darboux sums is
$$\begin{align}U(P,f) - L(P,f) &= \sum_{j=1}^m osc(f,\bar{I}_j) \,l(\bar{I}_j)+ \sum_{k=1}^n osc(f,\bar{J}_k) \,l(\bar{J}_k)\\ &\leqslant m 2|f(x_j)|\frac{\epsilon}{4m|f(x_j)|} + 2E(b-a) \\ &\leqslant \epsilon,\end{align}$$
where $osc(f,\bar{I}_j) = \sup_{x \in \bar{I}_j}f(x) - \inf_{x \in \bar{I}_j}f(x).$
Therefore, $f$ is integrable by the Riemann criterion.