Showing $F_n$ uniformly converges to $F$ if $f_n\to f$ and $F_n$ is the integral of $f_n$

Showing $F_n$ uniformly converges to $F$ if $f_n\to f$ and $F_n$ is the integral of $f_n$.

Is my thought process okay?

This is the problem:

Does it suffice to use the definition of uniform convergence, and that $\sup(F_n-F) = $ integral($f_n-f$) = integral($0$) as $f_n\to f$, so it converges?

Sorry for no latex.

Solution 1:

To write out your idea, it may go like the following. \begin{align*} \sup_{x\in[0,1]}\left|F_{n}(x)-\int_{0}^{x}f(t)dt\right|&=\sup_{x\in[0,1]}\left|\int_{0}^{x}(f_{n}(t)-f(t))dt\right|\\ &\leq\sup_{x\in[0,1]}\int_{0}^{x}|f_{n}(t)-f(t)|dt\\ &\leq\int_{0}^{1}\sup_{t\in[0,1]}|f_{n}(t)-f(t)|dt\\ &=\int_{0}^{1}\|f_{n}-f\|_{\text{sup}}dt\\ &=\|f_{n}-f\|_{\text{sup}}\\ &\rightarrow 0. \end{align*}