# Why is a random variable on a sigma algebra not necessarily measurable on a sub sigma algebra?

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?

• Consider the case $\mathcal{H} = \{\emptyset , \Omega \}$. Which functions are measurable here? Jan 17, 2022 at 23:50
• @JoseAvilez Would it only be only bernoulli random variables? Jan 17, 2022 at 23:55
• No. Only constants would be measurable. Jan 17, 2022 at 23:58
• @JoseAvilez So since this is a sub-sigma algebra of every sigma algbera on Omega, this is a counter example? Jan 17, 2022 at 23:59
• 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. Jan 18, 2022 at 0:02

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 $$E(X | \mathcal{G}) = Y$$.

• 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 Jan 18, 2022 at 1:24
• @SamKirkiles Pleasure! Good luck on your seminar course. 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.

• 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? Jan 18, 2022 at 1:14