We construct the desired step function for a given \(n \in \mathbb{N}\). Let

\begin{equation*} I_{n,j} = \bigg(-n + \frac{j}{2^n}, -n + \frac{j+1}{2^n}\bigg], \qquad \text{for } j = 1, \ldots, n2^{n+1} - 1 \end{equation*}

and \(I_{n,0} = \bigl[-n, -n + \frac{1}{2^n}\bigr]\) be a partition of the interval \([-n, n]\). Now choose \(b \in B\) and \(b_{n,j} \in \im f \cap I_{n,j}\) for \(j = 0, \ldots, n 2^n - 1\), if the intersection is non-empty, otherwise set \(b_{n,j} = b\).

Next, we define \(I'_{n,j} = f^{-1}(I_{n,j})\) and \(I'_{n} = f^{-1}([-n, n])\). These sets are \(\mathcal{A}\)-measurable, since for \(j = 0, \ldots, n 2^n - 1\) we have

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

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

\begin{equation*} f_n(\omega) = \sum_{j=0}^{n 2^n - 1} b_{n,j} 𝟙_{I'_{n,j}}(\omega) + b 𝟙_{(I'_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 \(f(\omega) \in \mathbb{R}\), there exists an \(n_0 \in \mathbb{N}\) such that \(f(\omega) \in [-n, n]\) for all \(n \ge n_0\). If we fix \(n \ge n_0\), there exists in particular a \(j \in \{0, \ldots, n 2^n - 1\}\) such that \(f(\omega) \in I_{n,j}\). Therefore, the intersection \(\im f \cap I_{n,j}\) is non-empty and consequently \(f_n(\omega) = b_{n,j} \in \im f \cap I_{n,j}\). From this, by construction,

\begin{equation*} |f_n(\omega) - f(\omega)| \le 2^{-n}, \end{equation*}

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