# Uniform convergence of a sequence of functions defined by an integral

I would like your opinion about the following question:

Let $\{f_n\}$ be a sequence of integrable functions in $[a,b]$which uniformly converges to integrable function $f$, then let $F(x)=\int_a^xf(t)dt$, and $F_n(x)=\int_a^xf_n(t)dt$, I want to prove that $F_n$ uniformly converges to $F$ on $[a,b]$.

Can I just use the continuous of the integral and say that $$\lim_{n \to \infty} F_n(x)=\lim_{n \to \infty}\int_{a}^{x}f_n(t)=\int_{a}^{x}\lim _{n \to \infty}f_n(t)dt=\int_{a}^{x}f(t)dt=F(x)?$$

Thanks a lot!

• Your argument is correct, but it only proves that $F_n$ converges pointwise to $F$. More work is needed to show that the convergence is uniform. – Julián Aguirre Feb 7 '12 at 9:01
• Thank you! Should I use the $\epsilon$-definition to show the uniformly convergence? I tried that one, with out much of successes. – Jozef Feb 7 '12 at 9:11

No, that shows convergence, but not uniform convergence. Fix $\epsilon>0$ and choose $N$, by uniform convergence of $(f_n)$, such that
$$n\geq N\Rightarrow |f(x)-f_n(x)|<\frac{\epsilon}{b-a} \text{for all } x$$
Then, for $x\in[a,b]$, $\int_a^x|f(y)-f_n(y)|dy<\epsilon$. Thus $|F_n(x)-F(x)|<\epsilon$. So $F_n$ converges uniformly to $F$ on $[a,b]$.