let $f:(0,1)$ be a non negative continuous function. show that $\int_{(0,1)} f$ exist

Question

let $f:(0,1)$ be a non negative continuous function. show that $$\int_{(0,1)} f$$ exist if and only if limit $\delta \to0$ $$\int _{\delta}^{1-\delta} f$$ exist.

How to approach this problem. Any hint ?

Answer 1

So we are dealing with one-sided limit $\delta\rightarrow 0^{+}$ to $\displaystyle\int_{\delta}^{1-\delta}f$. One-sided limit is equivalent to the sequential characterization of monotone sequence $(\delta_{n})$ such that $\delta_{n}>0$, $\delta_{n}\rightarrow 0$, in which case, $\delta_{n}\downarrow 0$, to that $\displaystyle\int_{\delta_{n}}^{1-\delta_{n}}f$, and this amounts to the Monotone Convergence Theorem because $f$ is nonnegative and $\displaystyle\int_{\delta_{n}}^{1-\delta_{n}}f=\int_{0}^{1}\chi_{[\delta_{n},1-\delta_{n}]}f$ and $\chi_{[\delta_{n},1-\delta_{n}]}f\leq\chi_{[\delta_{n+1},1-\delta_{n+1}]}f$, $n=1,2,...$ and $\chi_{[\delta_{n},1-\delta_{n}]}f\uparrow f$ on $(0,1)$.