= Solution
Since $f_n\uparrow f$, the functions
$$
g_n=f_n-f_1
$$
are nonnegative and increase to $g=f-f_1$. The monotone convergence theorem gives
$\mu(g_n)\uparrow\mu(g)$. Because $f_1$ is integrable, adding the finite number
$\mu(f_1)$ yields
$$
\boxed{\mu(f_n)\uparrow\mu(f)}.
$$
Integrability of $f_1$ is essential. On $\mathbb R$ with Lebesgue measure, let
$$
f_n=-\mathbf1_{[n,\infty)}.
$$
Then $f_n\uparrow0$ pointwise, but every integral is $-\infty$, so
$\mu(f_n)$ does not increase to $\mu(0)=0$.
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