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In the following question

Is a random variable constant iff it is trivial sigma-algebra-measurable?,

whose answer I copy here it is proven that

If $F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is a constant

If $f==c$ is constant it is ALWAYS measurable (for any sigma-algebra). This holds as $f^{-1}[C]$ is $X$ if $c \in C$ and empty if $c \notin > C$. And both sets are in any sigma-algebra.

On the other hand, if $f$ is $F$-measurable and non-constant, then it assumes at least two values $c_1$ and $c_2$. The set $f^{-1}[{c_1}]$ must be in $F$ (by being $F$-measurable, as ${c_1}$ is a closed set) but this set is non-empty (as $c_1$ IS a value of $f$) and not $X$ (as the points $x$ where $f$ assumes the value $c_2$ are not in it). So this set cannot be in $F$, and so $f$ must be constant.

However in my probability lecture this property was given like this

$F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is an a.e (almost everywhere) constant, f being a random variable

Is this a generalization? How do I prove it? I ended up proving it without the a.e, just as in the reference question, and I don't know how to deal with the a.e to make it more general. For instance in the converse: If $f=c$ a.e then there should be sets $N$ with null measure for wich the function is not equal to c, So when I take a borel set B and do the preimage $f^{-1}(B)$ what happens if N \subset B? this leads to the additional question of what is the preimage of a negligible set?

some_math_guy
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2 Answers2

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The almost-everywhere version is false: it's indeed the case that a function $f:X\to \Bbb R$ is constant if and only if it's $\{\emptyset,X\}$-measurable.

If you have a measure space $(X,\mathcal E,\mu)$ and a map $f:X\to \Bbb R$, it is true that $f$ is equal almost everywhere to a $\{\emptyset,X\}$-measurable function if and only if $f$ is almost everywhere constant, but it's hard to say if this was the intention of the claim, or if the statement you cite is just an error.

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    The context was the study of the conditional expectation of a random variable with repsect to a sub-sigma algebra. That is $\mathbb{E}[X|G]$ where G is a sub sigma algebra of F, the measure space is $(\Omega ,F,\mathbb{P})$, and we were analyzing the case when G was the trivial algebra – some_math_guy Dec 02 '20 at 23:02
  • Then she gave that property to conclude that if the conditional expectation is G-measurable, then it is a constant a.e Does this change anything? If not, how do you prove the version you are giving in the second paragraph? – some_math_guy Dec 02 '20 at 23:04
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    @mathlover Constant implies, by all means, constant almost everywhere, so this last assertion is fine. It's the other way around that fails. –  Dec 02 '20 at 23:06
  • The converse you mean? but she said it was an iff. Could you give a counterexample? – some_math_guy Dec 02 '20 at 23:08
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    @mathlover Of course I can. $\Omega$ is $\Bbb R$ with Lebesgue measure, $f=1_{{0}}$. It's constant almost-everywhere, but it isn't ${\emptyset,\Bbb R}$-measurable because $f^{-1}(-\infty,1/2]=\Bbb R\setminus{0}$. –  Dec 02 '20 at 23:10
  • @mathlover As for my second statement, it's tautological once you have the characterization of constant functions: a function is almost everywhere equal to a function which satifies a property that's equivalent to being constant, if and only if it is almost everywhere constant. –  Dec 02 '20 at 23:13
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    However, here the fact is that the "almost-everywhere" we use for conditional expectations is implicitly meant to be "$G$-almost-everywhere" (as opposed to what one could think of your post, i.e. that you mean almost-everywhere with respect to some given larger $\sigma$-algebra). The fact is that the notion of $G$-almost-everywhere trivialises to "everywhere" for $G={\emptyset,X}$: two functions are ${\emptyset,X}$-almost-everywhere equal if and only if they are equal. @mathlover –  Dec 02 '20 at 23:19
  • I am a bit confused with your counterexample. $\Omega$ is not $\mathbb{R}$ If I am not mistaken a random variable is a function $f: (\Omega, F,\mathbb{P}) \mapsto (\mathbb{R}, B ,m)$ where$ m$ is the Lebesgue measure. In your counterexample you are getting that the preimage is a subset of the real numbers, while in the context of random variables you should get a subset of the abstract $\Omega$ – some_math_guy Dec 02 '20 at 23:29
  • Can the measurable function you're giving be a random variable? – some_math_guy Dec 02 '20 at 23:30
  • @mathlover Sorry, let $\Omega=[0,1]$ with same measure, or $\Omega=\Bbb R$ and the measure being $A\mapsto\frac1{\pi}\int_A \frac1{x^2+1},dx$, or whatever lets you see the forest despite the trees. –  Dec 02 '20 at 23:31
  • So, is the a.e version still mistaken even if considering the "G-almost-everywhere" meaning? – some_math_guy Dec 02 '20 at 23:39
  • This answer is wrong without specifying the $\sigma$-algebra $\mathcal F$ on $\mathbb R$. For example, if $\mathcal F = {\emptyset, \mathbb R}$, then the identity function is measurable (and not constant). – James Dec 28 '22 at 05:48
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[This is a long comment]

$F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is an a.e (almost everywhere) constant, f being a random variable

We have $\Rightarrow$, but not $\Leftarrow$, as explained in the accepted answer. This question is settled.

However, I was thinking about what I believe to be the unspoken essence of your question, based on this comment. Here is how I'd phrase it:

For most practical purposes, we don't distinguish between a.e. equal variables. For example, between a constant $c$ and an a.e. constant $C\stackrel{\text{a.e.}}{=}c$. However, $c$ qualifies as a candidate for $E[X]$, but $C$ does not, because it is not $F$-measurable, unless $C\equiv c$. Why do we disqualify a.e. equal random variables from conditional expectation candidates?

More generally,

Why not defining random variables as equivalence classes?

The answer to the linked question is given therein: it has something to do with stochastic processes indexed by uncountable sets.

See also:

paperskilltrees
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