convergence of measurable function on Lebesgue measure I am trying to prove that $f_n$ converges to $0$ almost uniformly and in measure, but it does not converge to $0$ in $L^1 $. $f_n$ is defined as $f_n(x)=nI_{[0,1/n]}(x)$. If anyone can help with this problem, that will be very helpful.
It is clear that if $f_n$ converges uniformly to $f$, that means by definition, 
$|f_n(x)-f(x)| = |n| < \epsilon$ if and only if  $0 \leq x \leq \epsilon /n$. Can I conclude the result by letting $\epsilon =1$. but I am not sure this way is correct. Any comments on this will be a great help.
Thank you.

Let $\varepsilon>0$ and choose $N$ so that $\frac1N<\varepsilon$. Then $f_n(x)=0$ for $n\geqslant N$, so $f_n(x)=0$ for $x>\frac1n$. It is clear that $$\bigcap_{n=1}^\infty \left[0,\frac1n\right] = \{0\}, $$
so $$\lim_{n\to\infty} f_n(x) = \begin{cases}\infty,& x=0\\0,& x\ne 0.\end{cases} $$
Since $\mu(\{0\})=0$, we conclude that $f_n\to 0$ almost uniformly. Moreover,
$$\lim_{n\to\infty}\mu\left(\{x\in\left[0,\frac1n\right] :|f_n(x)|\geqslant\varepsilon \right)=0, $$
so $f_n\to 0$ in measure.
However, for any $n$, we have
$$\|f\|_1 = \int |f_n|\ \mathsf d\mu = n\mu\left(\left[0,\frac1n\right]\right) = 1, $$
so $f_n$ does not converge to $0$ in $L^1$.
