Hector Petersen

2022-06-20

A problem in Billingsley's Probability and Measure says that if $f:[0,1]\to \mathbb{R}$ 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

$f=0\text{on}[0,1/2)$

$f=1\text{on}[1/2,3/4)$

$f=2\text{on}[3/4,7/8)$

$\vdots $

$f=n\text{on}{\textstyle [}\frac{2n-1}{{2}^{n}},\frac{2n+1}{{2}^{n+1}}{\textstyle )}$

Then $f(x)<\mathrm{\infty}$ for all $x\in [0,1]$ 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?

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

$f=0\text{on}[0,1/2)$

$f=1\text{on}[1/2,3/4)$

$f=2\text{on}[3/4,7/8)$

$\vdots $

$f=n\text{on}{\textstyle [}\frac{2n-1}{{2}^{n}},\frac{2n+1}{{2}^{n+1}}{\textstyle )}$

Then $f(x)<\mathrm{\infty}$ for all $x\in [0,1]$ 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

Beginner2022-06-21Added 25 answers

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 $f:[0,1]\to \mathbb{R}$, then it is true for all Borel functions $f:[0,1]\to \mathbb{R}$."

"Show that if Lusin's theorem is true for all integrable [or bounded] functions $f:[0,1]\to \mathbb{R}$, then it is true for all Borel functions $f:[0,1]\to \mathbb{R}$."

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