 Solution

Since $f_n\uparrow f$, the functionsare 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

Integrability of $f_1$ is essential. On $\mathbb{R}$ with Lebesgue measure, letThen $f_n\uparrow 0$ pointwise, but every integral is $-\infty$, so
$\mu (f_n)$ does not increase to $\mu (0)=0$.