1.18

The exercise didn't specify anything about $f = \infty$ on sets with positive measure. I'll assume $f$ is a.e. finite. From here, it, of course, suffices to consider only non-negative measurable functions. For any such $f$, we need to find a sequence of continuous functions $\lbrace f_n \rbrace_{n = 1}^{\infty}$ such that the set: $$E = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} |f(x) - f_n(x)| > 0 \Big\rbrace$$ has measure zero. First, define the $d-$dimensional cubes: $$Q_n = [-n,n]\times \ldots \times[-n,n]$$ And the functions: $$g_n(x) = \min{\lbrace n,f(x) \rbrace}\chi_{Q_n}$$ It follows from Lusin's Theorem that for any of our $g_n$'s, we can find a set $F_n$ such that $g_n \chi_{F_n}$ is continuous, and $m(Q_n \backslash F_n)\leq \frac{1}{2^{n}}$. Now, we can simply define $f_n = g_n \chi_{F_n}$.

Lastly, define: $E_n = Q_n \backslash F_n$. Notice that: $$\sum_{n = 1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} \leq 1$$ Now, of course, we simply need to show that $E$ is $\sup{\lbrace E_n \rbrace}$. This actually isn't all that hard. First, notice: $$\bigcup_{n \geq k} E_n = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} |f(x) - f_n(x)| > 0, \hspace{0.25cm} n \geq k \Big\rbrace$$ Naturally, we arrive that the result that $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k} E_n$, and thus, by the Borel-Cantelli Lemma, $m(E) = 0$, as desired.