# I have somehow confused myself with this fairly straightforward proof. We need to show that &#x0

I have somehow confused myself with this fairly straightforward proof. We need to show that $\lambda f$ is a measurable function on $\left(S,\mathcal{S}\right)$, i.e. that for any $c\in \mathbb{R}:\left\{s\in S:f\left(s\right)\le c\right\}\subset S$. If $\lambda \ne 0$, then the claim follows immediately from the measurability of $f$, namely as $c/\lambda \in \mathbb{R}$ it follows that $\left\{s\in S:f\left(s\right)\le c/\lambda \right\}=\left\{s\in S:\lambda f\left(s\right)\le c\right\}$
But then, if $\lambda =0,\lambda f=0$ and I am not sure how to convince myself that $\lambda f$ is measurable.
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Aryanna Caldwell
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.
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aggierabz2006zw
$f:S\to \mathbb{R}$ measurable function.
$g:\mathbb{R}\to \mathbb{R}$ defined by $g\left(x\right)=\lambda x$ , $\lambda \in \mathbb{R}$ , is continuous.
Hence, $g\circ f=\lambda f$ is measurable.