Let $(X,\mathcal F,\mu)$ a measure space. Let $E\subset X$. Does saying "$E$ is $\mathcal F-$measurable" and "$E$ is $\mu-$measurable" are the same thing ? I would say no, and I would say that $E$ is $\mu-$measurable if for all set $A\subset X$, $$\mu^*(A)=\mu^*(E\cap A)+\mu^*(E^c\cap A),$$ and $E$ is $\mathcal F-$measurable if $E\in \mathcal F$. But in the same time I I don't see the interest to precise use for example caratheodory measurable set if we are in for example $(\mathbb R, \mathcal B, m)$ where $\mathcal B$ is the $\sigma -$algebra of Borel set and $m$ is the Lebesgue measure. So I don't really know what to think.

  • $\begingroup$ To the best of my knowledge, $\mathcal{F}$-measurable and $\mu$-measurable mean the same thing. $\endgroup$ – Yanko Jun 29 '18 at 14:39
  • $\begingroup$ @Yanko See here for instance. $\endgroup$ – drhab Jun 29 '18 at 14:40
  • $\begingroup$ Like you I would say no. When $(X,\mathcal F,\mu)$ is a measure space then a set $A$ can also be defined to be $\mu$-measurable if sets $B,C\in\mathcal F$ exist with $B\subseteq A\subseteq C$ and $\mu(C-B)=0$. $\endgroup$ – drhab Jun 29 '18 at 14:43

You're right that $\mu$-measurable is not the same thing as $\mathcal{F}$-measurable.

In the case when $\mu$ is Lebesgue measure, if you take $X = \mathbb{R}$ and $\mathcal{F} = \mathcal{B}$ (the Borel sets), then it is well known that (at least, assuming the axiom of choice) the $\sigma$-algebra of Lebesgue measurable sets is strictly larger than $\mathcal{B}$. This is covered in Real Analysis by Royden-Fitzpatrick, for example.

As for the meat of your question ("But in the same time..."), the simplest way to think of this is many of the operations we do yield Lebesgue measurable functions that may not be Borel measurable. For example, consider the space $L^{1}(\mathbb{R})$. It contains functions that are Lebesgue measurable, but not Borel measurable (e.g. characteristic function of a bounded set that is Lebesgue but not Borel measurable). On the other hand, $C^{\infty}_{c}(\mathbb{R})$ is dense in $L^{1}(\mathbb{R})$. (In fact, this is an underrated way of thinking about $L^{1}(\mathbb{R})$, as the completion of $C^{\infty}_{c}(\mathbb{R})$ with respect to the $L^{1}$-norm.) What this tells us is some of the functions we obtain in analysis (for example, as limits) will not be Borel measurable.

(Good but tedious exercise for the reader: what goes wrong if you try to prove that $L^{1}$-convergence of smooth functions implies the limit is Borel measurable.)

My understanding is issues regarding completeness of measure spaces (the $\mu$-measurable sets complete the space $(X,\mathcal{F},\mu)$; cf. Bogachev's Measure Theory) are even more important in stochastic processes, but I haven't worked with them enough yet to give more details.


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