# $\lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$ only when f is uniformly convergent?

$\lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$ .....(A)

where $f_1(x), f_2(x),......,f_n(x)$ represent a sequence of functions such that $f(x) = \lim_{n\rightarrow \infty} f_n(x)$

Is the statement (A) true only if $f(x)$ is uniformily convergent?

A sequence of functions f is said to be uniformly convergent in $[a,b]$ if $\forall ~ \epsilon>0, ~ \exists ~ m_o \in N: |f_n(x)-f(x)| < \epsilon , ~ \forall ~x \in [a,b] , ~ \forall ~ n \geq m_o$ i.e. $m_o$ is common for the convergence of $f(x_o) \forall x \in [a,b]$

So, a converging function whether uniformly convergent or NOT in the interval [a,b], will still follow the property that : $\lim_{n\rightarrow \infty} f_n(x)$ = $f(x) ~ \forall ~ x \in [a,b]$

Hence, shouldn't $\lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$ irrespective of whether f is uniformily convergent or not ?

• A sequence of functions $f_n$ is said to be uniformly convergent... And the correct order of the quantifiers is $\forall\epsilon>0~\exists m_0\in{\Bbb N}~ \forall~x\in [a,b]$... Commented Feb 3, 2014 at 17:53

The statement is not true in general.

Example: Take $f_n=n$ on $[0,\frac{1}{n}]$ and $0$ on $[\frac{1}{n},1]$. Then $\int\limits_{0}^{1}f_n=1$, but $f_n \rightarrow 0$.

Also, $\int f_n \rightarrow \int f$ doesn't imply uniform convergence.

e.g.: $f_n=x^n$ on $[0,1]$

Convergence of integrals does not imply uniform convergence of the functions. Consider $$f_n(x)=4x^n(1-x^n)$$ and $$f(x)=0$$ We have that $\lim\limits_{n\to\infty}f_n(x)=f(x)$ and $$\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=\int_0^1f(x)\,\mathrm{d}x$$ yet the convergence is not uniform since $\sup\limits_{[0,1]}f_n(x)=1$.

Convergence of the functions does not imply convergence of the integrals. Consider $$f_n=2nx^{n-1}(1-x^n)$$ and $$f(x)=0$$ We have that $\lim\limits_{n\to\infty}f_n(x)=f(x)$ yet $$\int_0^1f_n(x)\,\mathrm{d}x=1$$ and $$\int_0^1f(x)\,\mathrm{d}x=0$$

A classic example is $n x^n$ on $(0,1)$. The function converges pointwise to $0$, so

$$\int_0^1 \lim_n n x^n \,dx = 0$$

on the other hand $$\lim_n \int_0^1 n x^n \,dx = \lim_n \frac{n}{n+1} x^{n+1}\vert_0^1 = 1$$

The main idea is that an integral is concerned with global behavior of the function, and pointwise-only results are about as local as it gets.