On the continuity of a function given by the 'partial' Lebesgue integral of a multivariable function

Question

If somebody comes up with a better title, then please feel free to tell me.

Let $(\Omega,\mathcal F,\mu)$ be a measure space. Let $(a\,..b)\subseteq \mathbb R$ and $f : (a\,..b) \times \Omega \to \mathbb R$ be such that, $f(x,\_)$ (*) is $\mu$-integrable for all $x$ and $f(\_,\omega)$ is continuous for all $\omega$. Let $g : \Omega \to \mathbb R$ be $\mu$-integrable such that $|f(x,\_)| \leq g$ for all $x\in \Omega$.

Then one can show that $x\mapsto F(x):=\int f(x,\_) d\mu$ is continuous.

This was an exercise with a hint: "Apply dominated convergence theorem for a suitable sequence of functions."

My question is: How do I prove this? I'm completely stuck since I have no idea what kind of sequence I'm supposed to look at.

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(*) The symbol $\_$ is a placeholder for an argument, e.g. $f(x,\_)$ is a function $\Omega \to \mathbb R$.

Answer 1

Edit: changed hint to full answer (I'm assuming that $g$ is supposed to be $\mu$-integrable)

We want to show that $x \mapsto F(x)$ is a sequentially continuous map on $(a,b)$. Let $x_n \to x$ in $(a,b)$, define \begin{align} f_n : ~&\Omega \to \mathbb{R} \\ & \omega \mapsto f(x_n, \omega) \end{align} As $f(\cdot,\omega)$ is continuous for every $\omega \in \Omega$, we have $f(x_n,\omega) \to f(x,\omega)$ for every $\omega \in \Omega$, therefore $f_n$ converges pointwise to $f(x,\cdot)$. As $|f_n(\omega)| = |f(x_n,\omega)| \leq g(\omega)$, we may apply the dominated convergence theorem to the sequence $f_n$ on $L^1(\mu)$ to conclude

$$\lim_{n\to \infty} F(x_n) = \lim_{n\to\infty} \int_{\Omega}f_n ~ d\mu = \int_{\Omega} \lim_{n\to\infty} f_{n}~ d\mu = \int_{\Omega}f(x,\cdot)~d\mu = F(x)$$