 Hector Petersen

2022-06-20

A problem in Billingsley's Probability and Measure says that if $f:\left[0,1\right]\to \mathbb{R}$ is finite-valued and Borel-measurable, then $f$ may be assumed integrable, and even bounded.

I don't see how this is true though. For example define $f$ as

$⋮$

Then $f\left(x\right)<\mathrm{\infty }$ for all $x\in \left[0,1\right]$ and $f$ is measurable since it is a step function and takes only discrete values. However $f$ is not bounded on [0,1].
What's wrong with this attempted counterexample? How would the statement be proven? Marlee Guerra

As I suspected, you left out context. The problem is about proving Lusin's theorem for functions on [0,1]. When part (a) says "Show that $f$ may be assumed to be integrable [or bounded]", what Billingsley means is that
"Show that if Lusin's theorem is true for all integrable [or bounded] functions $f:\left[0,1\right]\to \mathbb{R}$, then it is true for all Borel functions $f:\left[0,1\right]\to \mathbb{R}$."

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