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I'm new to measure theory so I realize this is a simple question. I was reading: Essentially bounded function on $\mathbb{R}$. I see why the examples are essentially bounded, but not bounded.

Can we say that all bounded functions on $\mathbb{R}^n$ are essentially bounded measurable functions? Here I suppose measurable should be taken to mean measurable w.r.t. Lebesgue measure? Is that the usual measure to use on $\mathbb{R}^n$?

Moreover, does that mean that all bounded functions on $\mathbb{R}^n$ (or say $(0, \infty)^n)$ are in $L^\infty(\mathbb{R}^n)$?

To clarify, a function is on $L^\infty$ if it is bounded except on sets of zero measure. Should interpret this as we do not care what happens on sets of zero measure?

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  • $\begingroup$ Yes bounded functions are essentially bounded. $\endgroup$
    – zhw.
    Mar 29, 2020 at 19:02
  • $\begingroup$ Bounded functions are essentially bounded because the empty set has measure zero. $\endgroup$ Mar 29, 2020 at 19:06
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    $\begingroup$ Bounded functions need not be measurable, though: consider the indicator function of a non-measurable subset. $\endgroup$
    – user239203
    Mar 29, 2020 at 19:21

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That a function $f:\mathbb R^n\to\mathbb R$ is essentially bounded means there is some number $M\ge0$ for which the measure of the set $\{x\in\mathbb R^n : |f(x)|>M\}$ is $0.$

That a function $f:\mathbb R^n\to\mathbb R$ is bounded means there is some number $M\ge0$ for which the set $\{x\in\mathbb R^n : |f(x)|>M\}$ is empty.

Since the empty set has measure $0,$ but not every set of measure $0$ is empty, every bounded function is essentially bounded, but some essentially bounded functions are not bounded.

As for saying "we do not care," perhaps that depends on the context. One can say that the value of an integral of $f$ does not depend on what $f$ does on a set of measure $0,$ if if one cares only about the values of integrals, then one does not care about what happens on such sets.

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