# suppose $f_n\to f$ uniformly on [a,b]. Then $f$ is Riemann Integrable

I am trying to understand the proof of theorem 7.16 in Rudin's Principles of mathematical analysis.

Let $\alpha$ be monotonically increasing on [a,b]. Suppose $f_n\in{R}(\alpha)$ for n=1,2,3,... and suppose $f_n\to f$ uniformly on [a,b]. Then $f\in R(\alpha)$

Put $\epsilon_n=sup|f_n(x)-f(x)|$. then: $$f_n-\epsilon_n\leq f\leq f_n+\epsilon_n$$

so the upper and lower integral satisfy: $$\int_a^b(f_n-\epsilon_n)d\alpha \leq \int_{\_}fd\alpha \leq \int^{\_}fd\alpha\leq \int_a^b(f_n+\epsilon_n)d\alpha$$

I dont understand the above inequality. specifically, $\int_a^b(f_n-\epsilon_n)d\alpha \leq \int_{\_}fd\alpha$.

• If $f\leq g$, then $\int_- f d\alpha \leq \int_- g d\alpha$. – Rigel Dec 7 '17 at 19:10
• For standard Riemann integral, you can just use the fact that if all $f_n$ are continuous at some point, then so is their uniform limit, so if discontinuity sets of $f_n$ are all null, then so is the discontinuity set of $f$. I'm not sure what sort of integral this is here, but I assume the same principle works. – tomasz Dec 7 '17 at 19:18

Like Rigel pointed out in the comments, if $f\le g$ then their lower integrals satisfy the same inequality $\underline{\int}f\le\underline{\int}g$ by definition of the lower integral as the sup over all lower sums. Thus we immediately obtain $$\underline{\int}f_n-\epsilon_n\,d\alpha \le \underline{\int}f\,d\alpha.$$ Since $f_n-\epsilon_n$ is integrable, its lower integral is equal to its integral, so we obtain the inequality written in Rudin: $$\int f_n-\epsilon_n\,d\alpha \le \underline{\int}f\,d\alpha.$$