# Showing that $\displaystyle\underset{n\rightarrow \infty}{\lim}\int_0^1 f_n = \int_0^1\underset{n\rightarrow \infty}{\lim} f_n$

How to solve the following task:

Show that if $f_n$ is a sequence of uniformly converging mappings $f_n \in C[0,1]$, where $C[0,1]=\{f:[0,1]\rightarrow\mathbb{R} \;\mid\; f\; \text{continuous}\}$ then

$\displaystyle\underset{n\rightarrow \infty}{\lim}\int_0^1 f_n = \int_0^1\underset{n\rightarrow \infty}{\lim} f_n$.

How to tackle this problem? Should I consider using some metric function here or?

• Hi @ClementC. you're right. Let me double check my problem statement. – jjepsuomi Feb 9 '16 at 20:58
• I edited the question. – jjepsuomi Feb 9 '16 at 21:02

Write $$\left\lvert \int_0^1 f_n - \int_0^1 f\right\rvert \leq \int_0^1 \left\lvert f_n - f\right\rvert \leq \int_0^1 \lVert f_n - f\rVert_\infty = \lVert f_n - f\rVert_\infty.$$ What does uniform convergence give you then?