Let $(X,\mathcal{M},\mu)$ be a measure space and suppose $\{f_n\}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $\int f_1 \lt \infty$. Then $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$
Atempt:
Since $\{f_n\}$ are decreasing, and converges pointwise to $f$, then $\{-f_n\}$ is increasing pointwise to $f$. So by the monotone convergence theorem
$$ \int_X -f~d\mu = \lim_{n\to\infty}\int_X -f_n ~d\mu$$ and so $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$
Best Answer
The problem is that $-f_n$ increases to $-f$ which is not non-negative, so we can't apply directly to $-f_n$ the monotone convergence theorem. But if we take $g_n:=f_1-f_n$, then $\{g_n\}$ is an increasing sequence of non-negative measurable functions, which converges pointwise to $f_1-f$. Monotone convergence theorem yields: $$\lim_{n\to +\infty}\int_X (f_1-f_n)d\mu=\int_X\lim_{n\to +\infty} (f_1-f_n)d\mu=\int_X f_1d\mu-\int_X fd\mu$$ so $\lim_{n\to +\infty}\int_X f_nd\mu=\int_X fd\mu$.
Note that the fact that there is an integrable function in the sequence is primordial, indeed, if you take $X$ the real line, $\mathcal M$ its Borel $\sigma$-algebra and $\mu$ the Lebesgue measure, and $f_n(x)=\begin{cases} 1&\mbox{ if }x\geq n\\\ 0&\mbox{ otherwise} \end{cases}$ the sequence $f_n $ decreases to $0$ but $\int_{\mathbb R}f_nd\lambda =+\infty$ for all $n$.