A counter-example for how the Downward Monotone Convergence theorem can fail is given by; Let $n = 1$ so we are working in $\mathbb{R}^1$ and let $E_n = [n,\infty)$. Then $m(E_n) = \infty$ for all $n$. And $E_n$ satisfies the criterion $E_1 \supset E_2 \supset E_3 \supset \cdots$. However, $m(\cap_{n=1}^{\infty} E_n) = m(\emptyset) = 0 \neq \infty = \lim \, m(E_n)$.
So let's go over the proof of the theorem to see exactly where the finiteness of an $E_n$ is used. First, using the same idea from the first problem we can write each $E_n$ as a disjoint union:
$$
E_n = \cup_{k=n}^{\infty} F_n \, \bigcup \, S
$$
Where each $F_n$ is given by $F_n = E_n - E_{n+1}$, and $S$ is given by $S = \cap^{\infty} E_n$. It should be clear that these sets are mutually disjoint. Now the additivity of measures gives
$$
m(E_n) = m(S) + \Sigma_{k=n}^{\infty} m(F_k)
$$
The statement that $m(E_n)$ is finite for some $n$ is exactly the statement that $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ and that $m(S)$ is finite.
That $m(S)$ is finite is immediate from the monotonicity of measures and that $S \subset E_n$ for all $n$ along with that $E_n$ has finite measure for some $n$. And since the series $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ we have:
$$
\lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = 0.
$$
We get:
$$
\lim_n \, m(E_n) = m(S) + \lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = m(S).
$$
Best Answer
Take $f_n(x)=\frac{1}{n}\boldsymbol 1_{[0,n]}$. You have that $$\lim_{n\to \infty }f_n(x)=0,$$ but $$\lim_{n\to \infty }\int_{\mathbb R} f_n=1.$$