I have the following problem that says: If $(u_n)_{n=1}^\infty$ is a sequence in $\mathcal{L}^1$ which converges pointwise to 0, then $\lim_{n \rightarrow \infty }\int u_n \, d\mu=0$
My intuition is that this is FALSE. Because if I set $u_n=n\cdot 1_{[0,1/n]}$ the the integral is 1.
However, I am not if this actually meets the criteria of pointwise convergence
Any hint would be appreciated
On
You have that $$u_n(0)=n \quad \forall n \in \mathbb{N}$$ So $u_n(0) \to +\infty$, i.e. $u_n \not \to 0$ pointwise, so you just need to fix the problem at $0$.
Another example might be $u_n=1_{[n,n+1]}$ or $u_n=\frac{1}{2n}1_{[-n, n]}$. The second one is more interesting, because $u_n \rightrightarrows 0$ with $\int u_n=1$ shows that why the uniform convergence is not enough to interchange the limit and integral.
It doesn't actually, but almost. Let, instead, $u_n=n\cdot 1_{(0,1/n]}.$ Then, as you say, $\int u_n\textrm{d}\mu=1$ always, where $\mu$ is the one-dimensional Lesbegue measure. Now, let's prove pointwise convergence to $0$. Let $x\in\mathbb{R}$. Then, if $x=0,$ we have $u_n(x)=0$ for all $x$. Hence, $\lim_{n\to \infty}u_n(x)=0$. If, however, $x\neq 0,$ then for all $n>1/|x|,$ we have $x\not\in (0,1/n]$ and thus, $u_n(x)=0$. This implies $\lim_{n\to\infty} u_n(x)=0$. Thus, we have pointwise convergence.