Is $f_n$ guaranteed to have a pointwise limit? Let $f_n$ be a sequence of non-negative continuous functions on $[0,1]$ such that $\lim_{n\to \infty}\int _0^ 1 f_n(x) dx=0$. 
Is $f_n$ guaranteed to have a pointwise limit?
I think the answer is yes but can't prove it.Also I have not got any counter-examples also.Is it true?
 A: Let $f_n$ be the function defined as follows:


*

*For ${1\over n}<x\le 1$, $f(x)=0$.

*For $0\le x\le {1\over n}$, the graph of $f$ is a line connecting $(0, \sqrt{n})$ and $({1\over n}, 0)$.
Or, we can define it more symbolically as:
$$ f_n(x)=\begin{cases}n^{1\over 2}-xn^{3\over 2} &\text{ for }\quad 0\le x\le {1\over n}\\0 &\text { for }\quad {1\over n}<x\le 1\end{cases}$$
Then:


*

*Each $f_n$ is continuous - to see that the two parts of $f_n$ "glue together," note that $$\lim_{x\rightarrow {1\over n}^-}f_n(x)=\lim_{x\rightarrow{1\over n}}[n^{1\over 2}-xn^{3\over 2}]=n^{1\over 2}-n^{-1}n^{3\over 2}=n^{1\over 2}-n^{1\over 2}=0=\lim_{x\rightarrow{1\over n}^+}f_n(x);$$

*$\int_0^1f_n(x)dx={1\over 2\sqrt{n}}\rightarrow 0$; but

*$\lim_{n\rightarrow\infty}f_n(0)$ does not exist.
A: The most intuitive way of thinking about this is trying to identify a "function" $f$ that is infinity at a single point but equals 0 elsewhere and having the $f_n$ "converge" to $f$. To be more precise, let $\{f_n\}$ be a sequence of functions with $\int_0^1 f_n \, \mathrm{d}x = 1/n$ and $\max{f_n} = f(0) = n$. Then $\lim_{n \to \infty} f_n(0)$ doesn't exist so the $f_n$ do not converge pointwise to some function.
A: An alternative sequence of functions as counterexample:
$$f_n(x) = \begin{cases} x^n & \text{ if } n \text{ is even}\\
0 & \text{ otherwise}\end{cases} $$
Then $\int_{0}^1 f_{2n} = \frac{1}{2n+1}$ and $\int_{0}^1 f_{2n+1} = 0$, the $f_n$'s are continuous and non-negative, and yet $f_{2n}(1) = 1$, $f_{2n+1}(1) = 0$ for all $n$ (so that the limit of $f_n(1)$ does not exist). 
