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Since \(X\) is \(\sigma\)-compact, there exists a sequence of compact sets \((K_n)_{n \in \mathbb{N}}\) such that \(K_n \subset K_{n+1}\) for \(n \in \mathbb{N}\) and \(X = \bigcup_{n \in \mathbb{N}} K_n\).

We now construct the desired step function for a given \(n \in \mathbb{N}\). Since \(K_n\) is compact, there exist finitely many points \(x_j \in K_n\), \(j = 1, \ldots, k_n\), with

\begin{equation*} K_n \subset \bigcup_{j=1}^{k_n} B_{\frac{1}{n}}(x_j). \end{equation*}

Then there exist measurable pairwise disjoint sets \(M_{n,j} \subset X\) such that

\begin{equation*} K_n = \bigcup_{j=1}^{k_n} M_{n,j}. \end{equation*}

Now let \(b \in B\) and \(b_{n,j} \in B \cap M_{n,j}\) for \(j = 1, \ldots, k_n\), if the intersection is non-empty; otherwise, set \(b_{n,j} = b\). We define \(M'_{n,j} = f^{-1}(M_{n,j})\) and \(M'_{n} = f^{-1}(K_n)\). These sets are \(\mathcal{A}\)-measurable because for \(j = 1, \ldots, k_n\) we have

\begin{equation*} f^{-1}(M_{n,j}) = f^{-1}(M_{n,j}) \cap f^{-1}(B) = f^{-1}(M_{n,j} \cap B) \in \mathcal{A}, \end{equation*}

since \(f\) is \((\Omega, \mathcal{A})\)-\((B, \mathcal{B}(B))\) measurable. Similarly, it follows that \(M'_n \in \mathcal{A}\). Finally, we define the step function

\begin{equation*} f_n(\omega) = \sum_{j=1}^{k_n} b_{n,j} 𝟙_{M'_{n,j}}(\omega) + b 𝟙_{(M'_n)^\mathrm{c}}(\omega), \end{equation*}

which is clearly \((B, \mathcal{B}(B))\)-measurable.

It remains to show that \(f_n \to f\) pointwise. Let \(\omega \in \Omega\). Since \(X = \bigcup_{n \in \mathbb{N}} K_n\), there exists an \(n_0 \in \mathbb{N}\) such that \(f(\omega) \in K_n\) for all \(n \ge n_0\). Fixing \(n \ge n_0\), there exists a \(j \in \{1, \ldots, k_n\}\) such that \(f(\omega) \in M_{n,j}\). Hence, the intersection \(B \cap M_{n,j}\) is non-empty and therefore \(f_n(\omega) = b_{n,j} \in B \cap M_{n,j}\). By construction, it follows that

\begin{equation*} d(f_n(\omega), f(\omega)) \le \frac{1}{n}, \end{equation*}

and thus \(f_n(\omega) \to f(\omega)\) as \(n \to \infty\).