Integration wrt increasing functions and measures

Question

Suppose that $\mu_n$ and $\mu$ are measures on $(X, \Omega)$ such that $\mu_n$ increases to $\mu$. If $f_n$ and $f$ are nonnegative measurable functions such that $f_n$ increases to $f$, then $\int_X f_n \, d\mu_n$ converges to $\int_X f \, d\mu.$

By MCT, $\lim_{n \to \infty} \int_X f_n \, d\mu_n = \int_X f \, d\mu_n$. So it seems that I should show that $\int_X f \, d\mu_n = \int_X f \, d\mu$. This seems to be not a fruitful one, if not false.

Do you have an idea? Or a counterexample?

Edit: I found this. The accepted answer uses Radon-Nikodym. Can we prove it without R-N?

Answer 1

The monotone convergence theorem helps, but it is a result for a fixed measure, though the Radon-Nikodym theorem makes it possible to reduce to that case. Here is a more elementary argument:

First, note that for all $n$, $$\int f_n \, \mathrm d\mu_n \leq \int f_n \, \mathrm d\mu\leq \int f \, \mathrm d\mu, $$ so $\int f \, \mathrm d\mu$ is an upper bound of the increasing sequence $(\int f_n \, \mathrm d\mu_n)$. It suffices to show it is the least upper bound. Let's assume that $\int f \, \mathrm d\mu<\infty$, the argument for the other case is similar. Let $\epsilon>0$. By monotone convergence, there is some $n$ such that
$$\int f \, \mathrm d\mu-\int f_n \, \mathrm d\mu<\epsilon/3.$$ Moreover, there must be a simple function $g\leq f_n$ such that $$\int f_n \, \mathrm d\mu-\int g \, \mathrm d\mu<\epsilon/3.$$ Finally, the set-wise convergence of the measures and $g$ being simple implies that for some $l\geq n$, we have $$\int g \, \mathrm d\mu-\int g \, \mathrm d\mu_l<\epsilon/3.$$ Together, $$\int f_l \, \mathrm d\mu_l\geq \int f_n \, \mathrm d\mu_l\geq \int g \, \mathrm d\mu_l\geq\int g \, \mathrm d\mu-\epsilon/3\geq \int f_n \, \mathrm d\mu-2\epsilon/3\geq \int f \, \mathrm d\mu-\epsilon.$$