which of the following statements are true or false? let $f_n(x)$, for $n\ge 1$, be a sequence of continuous non negative functions on $[0,1]$, such that
$$\ \lim_{n\to \infty} \int_0^1 f_n(x) \, dx=0.$$
which of the following statements is always correct?
A. $f_n\to 0$ uniformly on $[0,1]$
b. $f_n$ may not converge uniformly but converges pointwise.
c. $f_n$ will converge pointwise and limit may be non zero.
d. $f_n$ is not guaranteed to have a pointwise limit.
can anyone help me to solve above problem.
i know that if $f_n$ converges uniformly then limit of interal converges to interal of limit.
 A: HINT:
If $f_n(x)=(1-x)^n$, then 
$$\int_0^1 (1-x)^n\,dx =\frac{1}{n+1}\to 0 \,\,\text{as}\,\,n\to \infty$$
But 
$$\lim_{n\to \infty }f_n(x)=\begin{cases}1&,x=0\\\\0&,1\ge x>0\end{cases}$$
and therefore, the convergence of $f_n(x)$ is clearly not uniform.
SPOLIER ALERT Scroll over the highlighted area to reveal the solution

If $f_n(x)=\sin(nx)g(x)$,where $g(x)$ is continuously differentiable, then it is easy to show that  $$\int_0^1 f_n(x)\,dx\to 0$$(use the Riemann Lebesgue Lemma or integrate by parts).  Noting that $\lim f_n(x)$ fails to exist for $x\ne 0$ and we conclude the answer is (d) $f_n$ is not guaranteed to have a pointwise limit.

A: The answer is (d) in case you still can not see it by considering the function 

$$  \sin(nx) . $$

A: Hint 
Take for the sequence of functions $(f_n)$ a hill which is  getting narrower, rolling around and around in the interval $(0,1)$ and such that it's integral is equal to a sequence of reals converging to $0$. I let you write a precise case of such a sequence of functions.
$(f_n)$ doesn't converge pointwise proving that all four assumptions are wrong.
