I have a question regarding the following bolded claim made on Wikipedia

Consider the following:

  • $(\Omega, \mathcal{F}, P)$ is a probability space.
  • $X: \Omega \rightarrow \mathbb{R}^{n}$ is a random variable on that probability space with finite expectation.
  • $\mathcal{H} \subseteq \mathcal{F}$ is a sub-\sigma-algebra of $\mathcal{F}$.

Since $\mathcal{H}$ is a sub $\sigma$-algebra of $\mathcal{F}$, the function $X: \Omega \rightarrow \mathbb{R}^{n}$ is usually not $\mathcal{H}$-measurable, thus the existence of the integrals of the form $\left.\int_{H} X d P\right|_{\mathcal{H}}$, where $H \in \mathcal{H}$ and $\left.P\right|_{\mathcal{H}}$ is the restriction of $P$ to $\mathcal{H}$, cannot be stated in general. However, the local averages $\int_{H} X d P$ can be recovered in $\left(\Omega, \mathcal{H},\left.P\right|_{\mathcal{H}}\right)$ with the help of the conditional expectation. A conditional expectation of $X$ given $\mathcal{H}$, denoted as $\mathrm{E}(X \mid \mathcal{H})$, is any $\mathcal{H}$ measurable function $\Omega \rightarrow \mathbb{R}^{n}$ which satisfies: $$ \int_{H} \mathrm{E}(X \mid \mathcal{H}) \mathrm{d} P=\int_{H} X \mathrm{~d} P $$

Why is a random variable on a sigma algebra not necessarily measurable on a sub sigma algebra? If a sub-sigma algebra is a collection of subsets from the original sigma algebra which forms a sigma algebra in its own right, why wouldn't this enable any random variable to continue to be measurable on some sub-sigma algebra?

  • $\begingroup$ Consider the case $\mathcal{H} = \{\emptyset , \Omega \}$. Which functions are measurable here? $\endgroup$ Jan 17, 2022 at 23:50
  • $\begingroup$ @JoseAvilez Would it only be only bernoulli random variables? $\endgroup$ Jan 17, 2022 at 23:55
  • $\begingroup$ No. Only constants would be measurable. $\endgroup$ Jan 17, 2022 at 23:58
  • $\begingroup$ @JoseAvilez So since this is a sub-sigma algebra of every sigma algbera on Omega, this is a counter example? $\endgroup$ Jan 17, 2022 at 23:59
  • $\begingroup$ It's an example that exhibits why an $\mathcal{F}$-random variable need not be measurable with respect to $\mathcal{H}$. If $X$ is any non-constant random variable that is $\mathcal{F}$-measurable, it cannot be $\mathcal{H}$-measurable. $\endgroup$ Jan 18, 2022 at 0:02

2 Answers 2


The sequence of comments above was getting a bit long, so I'll convert it into an answer.

As noted in @William's answer, making the domain's $\sigma$-algebra smaller while keeping the co-domain's $\sigma$-algebra fixed makes it harder for a function to be measurable.

As an example, if we take any non-constant random variable $X$ on a $\sigma$-algebra $\mathcal{F}$, and then set $\mathcal{G} = \{\emptyset, \Omega\} \subset \mathcal{F}$, then we note that $X$ cannot be measurable, as only constants are $\mathcal{G}$-measurable.

As an example for computing a conditional expectation, set $\Omega = [0,1]$, $\mathcal{F} = \mathcal{B}[0,1]$, and $P = \lambda$ (Lebesgue measure). We now set $\mathcal{G} = \{\emptyset , [0,1], [0,0.5), [0.5, 1]\}$ and relativise $\lambda$ to $\mathcal{G}$ (i.e. we assign Lebesgue measure to the sets in $\mathcal{G}$). Consider the random variable $X : \Omega \to \mathbb{R}$ given by $X(\omega ) = \omega$. This is $\mathcal{F}$-measurable, but it is not $\mathcal{G}$-measurable.

Notice that $\int_{[0,0.5)} X dP = \frac{1}{8}$ and $\int_{[0.5, 1]} X dP = \frac{3}{8}$. We now wish to find a $\mathcal{G}$-measurable function that integrates to these same values. For a function to be $\mathcal{G}$-measurable, it must be constant on $[0,0.5)$ and $[0.5,1]$. Thus, we may set $$Y(\omega) = \begin{cases} \frac{1}{4} & 0 \leq \omega < 0.5 \\ \frac{3}{4} & 0.5 \leq \omega \leq 1 \end{cases}$$ Notice that $Y$ is $\mathcal{G}$-measurable and $\int_{[0,0.5)} Y dP = \frac{1}{8}$ and $\int_{[0.5, 1]} Y dP = \frac{3}{8}$. Thus $E(X | \mathcal{G}) = Y$.

  • $\begingroup$ Thank you for your many comments and very instructive guidance! This helps so much for preparing for my seminar course which uses conditional expectation as assumed knowledge $\endgroup$ Jan 18, 2022 at 1:24
  • $\begingroup$ @SamKirkiles Pleasure! Good luck on your seminar course. $\endgroup$ Jan 18, 2022 at 1:25

A function $f : (X_1,\Sigma_1,\mu_1) \to (X_2,\Sigma_2,\mu_2)$ is $(\Sigma_1,\Sigma_2)$-measurable if for all $M \in \Sigma_2$, it holds that $f^{-1}(M) \in \Sigma_1$.

If $S \subset \Sigma_1$, it may hold that $f^{-1}(M)$ is not in it, which is the technical statement of the bolded text.

As a general example, suppose that $f$ is $(\Sigma_1,\Sigma_2)$-measurable, and $\varnothing \neq M \subsetneq X_2$ so that $M \in \Sigma_2$. If $S$ is any $\sigma$-subalgebra of $\Sigma_1$ which does not contain $f^{-1}(S)$, then $f$ is not $(S,\Sigma_2)$-measurable.

As a concrete example, if the trivial $\sigma$-algebra $\{\varnothing,X_1\}$ is $T$, then any nonconstant which is $(\Sigma_1,\Sigma_2)$-measurable ($\Sigma_1 \neq T$) is not $(T,\Sigma_2)$-measurable.

A high-level takeaway of this fact is that although the codomain's $\sigma$-algebra plays a vital role in measurability, it is the domain's $\sigma$-algebra that can "make it hard" to be measurable. Indeed, see that for a fixed codomain with a fixed $\sigma$-algebra, the fewer sets in the domain's $\sigma$-algebra, the harder it is for a function between those spaces to be measurable with respect to those $\sigma$-algebras.

  • $\begingroup$ If someone were to say to compute the conditional expected value with respect to some set, would this mean to produce a random variable? Or to produce the expected value of that random variable? $\endgroup$ Jan 18, 2022 at 1:14

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