Showing that if $\lambda \in \mathbb{R}$ and $f$ is a measurable real valued function on $(S, \mathcal{S})$, then so is $\lambda f$ I have somehow confused myself with this fairly straightforward proof. We need to show that $\lambda f$ is a measurable function on $(S, \mathcal{S})$, i.e. that for any $c \in \mathbb{R}: \{s \in S: f(s) \leq c\} \subset S$. If $\lambda \neq 0$, then the claim follows immediately from the measurability of $f$, namely as $c/\lambda \in \mathbb{R}$ it follows that $\{s \in S: f(s) \leq c/\lambda \} = \{s \in S: \lambda f(s) \leq c \}$
But then, if $\lambda = 0, \lambda f = 0$ and I am not sure how to convince myself that $\lambda f$ is measurable.

If $\lambda=0$, then your function $\lambda f=0$ is identically $0$. The inverse image of any measurable set containing $0$ is $S$ and of any set not containing $0$ is the empty set. Both of these are measurable.
