# Converging sequence of integrals for uniformly bounded functions

Let ${f_n}$ be a uniformly bounded sequence of measurable functions on $[0,1]$. Define $$F_n(x)=\int_0^x f_n(t)dt.$$ Then there is a subsequence $\{F_{n_k}\}$ that converges uniformly on $[0,1].$

Attempt: We know $|f_n| \leq M$ for all $n \in \mathbb{N}$ with $M$ the least such upper bound. So $|F_n| \leq M(1-0)=M$ for all $n \in \mathbb{N}$ and $x \in [0,1]$. Since $\{|F_n|\}$ is a bounded and closed sequence of real numbers, we know there is a subsequence that converges to $\limsup |F_n|$.

We also know that, given $k \in \mathbb{N},$ there is a $j \in \mathbb{N}$ such that $||F_{n_l}|-\limsup |F_n|| \leq \frac{1}{k}$ for all $l \geq j$ due to a property of $\limsup.$ But this says exactly that there is a subsequence of $|F_n|$ that converges uniformly on $[0,1]$. $\square$

I know this answer is incomplete by a long shot (I need to look at the functions, and I'm assuming I'm taking the entire integral over $[0,1]$ in the proof), but I would like to know if the approach is hinting on anything, and regardless, what I need to do from here. Thanks!

• You mean $f_n(t)$ as an integrand. Aug 9 '14 at 1:27

For a fixed bounded function $f(x)$, $\int_0^xf(t)dt$ is uniformly continuous in $x$. Show that $F_n(x)$ is equicontinuous and uniformly bounded on the unit interval. Now apply Arzela Ascoli.