Can you help me solve this?
Let $f$ be a continuous, monotonically increasing function on $\Bbb R$. Let $g$ be a non-negative measurable function over a measurable set $\Omega$. Prove that:
$$\lim_{n\to \infty} \int_{\Omega}f\left(\left(1-\frac{1}n\right)g\right) = \int_{\Omega}f(g).$$
I used Beppo Levi's theorem on functions $G_n(x)= F_n(x)- F_1(x)$ where $F_n(x)=f((1-\frac{1}n)g(x))$ and $F_1(x)=f(0)$.
My problem is that I don't have any hypothesis on the integrability of $f$ nor a condition on the measure of $\Omega$.