Does any bounded continuous function preserve $L^2$ convergence?

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How to prove the following statement:

Given $f\in C(\mathbb{R})$ which is bounded and $\Omega\subset\mathbb{R}^n$ is a bounded domain, then for any $u\in L^2(\Omega)$, we have $f(u)\in L^2(\Omega)$ and $f$ is a continuous mapping from $L^2(\Omega)$ to $L^2(\Omega)$ in the sense that

$u\to f\circ u$ is continuous on $L^2(\Omega)$

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Due to the boundedness assumptions, $f(u)$ is in $L^p(\Omega)$ for all $p\in[1,+\infty]$.

Let $(u_n)$ converge to $u$ in $L^2(\Omega)$. Choose pointwise converging subsequence $(u_{n_k})$. Then $f(u_{n_k})$ converges pointwise to $f(u)$. In addition we have the integrable upper bound $$ |f(u)(x) - f(u_{n_k})(x)|^2 \le 4M^2, $$ convergence $f(u_{n_k})\to f(u)$ follows by dominated convergence theorem. Since the limit is independent of the taken subsequence, $f(u_n)\to f(u)$ in $L^2$ follows.