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Reference

This question is related to: Banach Spaces: Uniform Integral vs. Riemann Integral

Problem

What are examples of real-valued functions:

Bounded & Non-Step & Non-Measurable

(Especially, it should be not a.e. a step!)

C-star-W-star
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2 Answers2

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Let $V$ be a non-measurable set, it must be uncountable, select a countable points from it $\{x_i\}$, define $f=\chi_V$ except those points, and let $f(x_i)=\frac{1}{n}$.

So $f$ is bounded by $1$, unmeasurable, has countable many values.

John
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  • That is still to simple as you can subtract a bounded measurable function from it to get it piecewise-constant. – C-star-W-star Nov 21 '14 at 17:42
  • If singletons are measurable then the one restricted to the countable many points. – C-star-W-star Nov 21 '14 at 17:46
  • @JohnZHANG You can add a function which is $1-1/n$ at $x_n$ and $0$ elsewhere to get back an indicator function. Such a function is measurable, since its support has measure zero and the Lebesgue measure is complete. – Ian Nov 21 '14 at 17:46
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It is still quite trivial but at least...

Given the Lebesgue measure $\lambda:[0,1]\to(0,\infty)$.

Consider a Vitali set $\mu_*(V)<\mu^*(V)$.

Construct the function: $$f:[0,1]\to\mathbb{R}:f(x):=x^2\chi_V(x)$$ This one is not sum of measurable plus step but product.

C-star-W-star
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