Integrating over a sequence of sets $A_n$ with $\mu(A_n)\to0$ I am going through the proof of the following.

Let $(X,\mu)$ be a measure space and $f\colon X\to\overline{\mathbb R}$ be a measurable function with finite integral. If $A_1,A_2,\dots$ are $\mu$-measurable and $\lim_{n\to\infty}\mu(A_n)=0$, then $$\lim_{n\to\infty}\int_{A_n}f\,d\mu=0.$$

Proof.

Let $E_N=\{x\mid f(x)\leq N\}$ and define $f_N=\chi_{E_N}f$. Clearly
  $f_N\leq f$ and since $f$ has finite integral it is finite a.e., and
  it follows that $\lim_{N\to\infty} f_N=f$ a.e.; then,
  $\lim_{N\to\infty}(f-f_N)=0$ a.e.. Now $f-f_N$ is dominated by
  $|f|+|f_N|$ which has finite integral by additivity and hence by the
  DCT we get $\int (f-f_N)\,d\mu\to0$ as $N\to\infty$.
Now by additivity and the fact that $f_N\leq N$ for all $N$ we have 
  $$\int_{A_n}f\,d\mu=\int_{A_n}(f-f_N)\,d\mu+\int_{A_n} f_N\,d\mu\leq
 \int_{A_n}(f-f_N)\,d\mu+N\mu(A_n)$$
Taking $N\to\infty$ and noting that $\int_{A_n}(f-f_N)\,d\mu\leq
\int(f-f_N)\, d\mu$ gives $$\int_{A_n}f\,d\mu\leq 0+\infty\cdot\mu(A_n).$$ Taking $n\to\infty$ gives the result because $\infty\cdot 0=0$. 

Now I don't really see how the last line proves anything. How do we know that $f$ is not negative? As I see, we have only shown that the limit is not positive. It also makes me a little uncomfortable that we write $\infty\cdot\mu(A_n)$, I mean I know that in the extended reals $0\cdot \infty=0$ but is this really rigorous?
Any ideas? 
 A: I agree with @JohnMa that the proof is not rigorous since, by definition,
$$\infty \cdot \mu(A_n) = \begin{cases} \infty, & \mu(A_n)>0, \\ 0, & \mu(A_n) \end{cases}$$
and so the estimate just shows $\int_{A_n} f \, d\mu \leq \infty$.

Without loss of generality, we may assume $f \geq 0$ (otherwise replace $f$ by $|f|$). Fix $R>0$, then
$$\int_{A_n} f \, d\mu = \int_{A_n \cap \{f \leq R\}} f \, d\mu + \int_{A_n \cap \{f>R\}} f \, d\mu$$
implies
$$\int_{A_n} f \, d\mu \leq R \mu(A_n) + \int_{\{f>R\}} f \, d\mu.$$
Hence,
$$\limsup_{n \to \infty} \int_{A_n} f \, d\mu \leq \int_{\{f>R\}} f \, d\mu$$
for any $R>0$. Letting $R \to \infty$, it follows from the dominated convergence theorem (or monotone convergence theorem) that
$$\limsup_{n \to \infty} \int_{A_n} f \, d\mu \leq 0.$$
Since $f$ is non-negative, this shows
$$\lim_{n \to \infty} \int_{A_n} f \, d\mu = 0.$$
A: Following the notation of your proof, set $E_N = \{ x \in X \: \colon \vert f(x) \vert \leq N\}$ and $f_N = f \cdot \chi_{E_N}$ for all $N \in \mathbb{N}$. Then $\lim_{N \to \infty} f_N = f$ almost everywhere and $f_N \leq f$ for all $N$. Since $f$ is integrable, we can apply the DCT and find
\begin{equation}
\lim_{N \to \infty} \int \big| f - f_N \big| \, \textrm{d} \mu = 0.
\end{equation} 
Now, using that $\vert f_N \vert \leq N$, we find that
\begin{equation}\label{eq1}
\Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg|\leq \int_{A_n} \big| f - f_N \big| \, \textrm{d}\mu + N \mu(A_n) \leq \int \big| f-f_N \big| \, \textrm{d} \mu + N \mu(A_n).
\end{equation}
As $\mu (A_n)$ tends to $0$ as $n \to \infty$, we obtain
\begin{equation}
\limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \leq \int \big| f - f_N \big| \, \textrm{d}\mu + N \cdot 0 = \int \big| f - f_N \big| \, \textrm{d}\mu.
\end{equation}
This inequality holds for every $N \in \mathbb{N}$; in particular, it remains true in the limit $N \to \infty$, and hence:
\begin{equation}
\limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \leq 0.
\end{equation}
Thus, 
\begin{equation}
0 \geq \limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \geq \liminf_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \geq 0.
\end{equation}
This proves the claim.
