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my question is why $f$ measurable on $[0,\infty]$ implies $\{x: f(x+k)0$ and each $a \in \mathbb R$.

I ask because in the proof that f measurable and differentiable implies f' is measurable, we write $f'= \lim n\to \infty \frac{f(x+1/n)-f(x)}{1/n}$ and then since f(x) and $f(x+1/n)$ are measurable and the limit exists as f differentiable, then the limit is also differentiable.

But how do we see that f(x+1/n) is measurable? I tried to investiage the set $\{x: f(x+1/n)

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    Maybe we can use the fact that a translate of a measurable set is measurable? $$\{x:f(x+k)\lt a\}=\{x-k:f(x)\lt a\}$$ – 2017-02-20
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    The function $x \mapsto x+k$ is continuous for any $k$. – 2017-02-20
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    Oh thanks, so the function g that takes x to x+k is continuous and so f(g(.)) is measurable since f is measurable and g continuous? – 2017-02-20
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    -bof,thanks for the reply, but how to show that translates of that form are measurable? I know how to show for example if A measurable then A-k measurable, but here the conditions of the inclusion of each set is different, so its kind of confusing me whether we can use that fact without any further proof. – 2017-02-20

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