Let $(\Omega,\mathcal F,\mu)$ be a measure space and let $(X,\tau)$ be a topological space with countable base. Suppose we are given a function $f:\Omega \times X \to [0,\infty]$ with the following properties:
(i) For each $x\in X$, the map $\omega\mapsto f(\omega,x)$ is $\mathcal F$-measurable.
(ii) For $\mu$-almost every $\omega\in \Omega$, the map $x\mapsto f(\omega,x)$ is lower semicontinuous.
The exercise asks to show that the map $x\mapsto \int f(\omega,x) \, \mu(d\omega)$ is lower semicontinuous.
Attempt:
Since $X$ is first countable, lower semicontinuity of a map $h:X\to [-\infty,\infty]$ at $x_0 \in X$ is equivalent to the following statement: $h(x_0)\leq\liminf_{n\to \infty}h(x_n)$ whenever $(x_n)$ is a sequence in $X$ converging to $x_0$ (see here).
Fix $x_0\in X$ and let $(x_n)$ be a sequence in $X$ converging to $x_0$. Introduce the sequence of functions $(f_n)_{n\geq 0}$ defined by $f_n(\omega)=f(\omega,x_n)$. Note that each $f_n$ is measurable by (i). Now Fatou's lemma gives us $$\int f_0 \, d\mu\leq\int \liminf_{n\to \infty}f_n d\mu \leq \liminf_{n\to \infty} \int f_n\,d\mu $$
and since the sequence $(x_n)$ in $X$ converging to $x_0$ was arbitrary the result follows.
Is this correct? Thanks a lot for your help.