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Let f be defined on $\mathbb{R}^d,$ suppose $\int_{\mathbb{R}^d} |f(x)|\,dx<\infty$. Is f then measurable?

My question boils down to "Do I have to check that a given function is measurable, or can I suppose the function is measurable, compute the integral, and if the integral is finite, observe that the function was indeed measurable?"

manofbear
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    How do you compute the integral of a function that is not measurable? – mathcounterexamples.net Sep 26 '15 at 21:48
  • I suppose you can't. Which is why I was wondering: If you suppose the function is measurable and integrate as usual, and the integral of |f| is finite, does this mean f was measurable?

    The idea: Without knowing f is measurable, $\int |f|,dx$ is not defined. If we suppose it is defined and we get something finite, have I shown that the integral was defined in the first place, or is my result meaningless?

     – manofbear Sep 26 '15 at 21:59
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    See this question . – Tony Piccolo Sep 27 '15 at 05:39
  • it can happen that $|f|$ is measurable even if $f$ is not! By slightly modifying this construction, we can even have $|f|$ integrable without $f$ being measurable. – PhoemueX Sep 27 '15 at 08:22
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    Let $E\subseteq [0,1]$. Suppose $E$ is NOT Lebesgue measurable. Let $f$ be defined on $\mathbb R$ as $f= \chi_{[0,1]-E} - \chi_E$ (where $\chi_{[0,1]-E}$ and $\chi_E$ are the indicator functions of $[0,1]-E$ and $E$ respectively). Then $|f|=\chi_{[0,1]}$ and $\int_{\mathbb R} |f(x)|,dx = 1 <\infty$. But $f$ is NOT measurable. – Ramiro Sep 27 '15 at 18:14

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