# Why does this argument for conditional expectation fail?

I have a question and I know it is wrong. However I do not understand where I am messing up. If somebody could explain where I am going wrong, that would be great.

If we have a probability space $$(X,\mathcal{F},\mu)$$, and a sub-sigma-algebra $$\mathcal{A}\subset \mathcal{F}$$, the conditional expectation for $$f\in L^1(X,\mathcal{F},\mu)$$ is defined to be the (a.s) unique measurable $$\mathbb{E}[f\mid \mathcal{A}]\in L^1(X,\mathcal{A},\mu)$$ such that $$\int_A f\,d\mu=\int_A \mathbb{E}[f\mid \mathcal{A}]\,d\mu$$ for all $$A\in \mathcal{A}$$.

My question is does $$\int_X f\,d\mu$$ not satisfy this? Of course, it would be nonsensical to say that the conditional expectation wrt some arbitrary sub-sigma-algebra is equal to the expectation.

However, if we plug $$\int_Xf\,d\mu$$ into the above for $$\mathbb{E}[f\mid \mathcal{A}]$$, we have $$\int_A(\int_Xf\,d\mu)\,d\mu=\int_X(\int_A fd\mu)d\mu=\int_Afd\mu$$ where the first equality uses Fubini.

I think the use of Fubini is suspicious. Any further explanations are welcome!

(INFO: The reason I came across this was as follows. I was looking at a proof of Birkhoff's Ergodic theorem, and in it it claimed that the conditional expectation with respect to the sigma-algebra consisting of sets of measure $$0$$ or $$1$$ was equal to the expectation. I believe this uses Fubini?)

EDIT: Sorry if I did not explain it correctly. I basically am saying that the constant function $$\int_X fd\mu$$ is in $$L^1(X,\mathcal{A},\mu)$$ and satisfies $$\int_A fd\mu=\int_A (\int_X fd\mu)d\mu$$, and so does this not mean $$\int_X fd\mu= \mathbb{E}[f\mid \mathcal{A}]$$?

• Perhaps I'm misunderstanding, but aren't you just conditioning with respect to $\mathcal{F}$ itself (that is, not really doing much at all)? – user296602 Apr 5 at 23:49
• Your calculation involves applying Fubini's theorem, on the product space $A\times X$, to a function that you write as $f$. There are two functions on $A\times X$ that might be called $f$. One sends each point $(a,x)$ to $f(a)$; the other sends each $(a,x)$ to $f(x)$. It seems to me that you may have switched from one of these meanings of $f$ to the other in the middle of your Fubini calculation. – Andreas Blass Apr 6 at 0:29
• Fubini's theorem is about exchanging the order of integrations of a function of two variables. What is exactly the two-variable function here? Here, it might help to see things clearly if you use the more explicit notation $\int g(x) d\mu(x)$ instead of $\int g d\mu$. – Blackbird Apr 6 at 12:55
• Thanks Blackbird. I thought for example F(x,a)=f(x) may work, but by doing this a $\mu(A)$ crops up (as one would expect!). – Mr Martingale Apr 6 at 13:38

## 1 Answer

$$E(f|\mathcal A)$$ is an $$\mathcal A$$ measurable random variable $$g$$ such that $$\int_A f d\mu =\int_A g d\mu$$ for all $$A \in \mathcal A$$. You want to claim that the constant $$g=\int_X f d\mu$$ satisfies these two properties. The measurability part is fine but the equation $$\int_A f d\mu =\int_A g d\mu$$ is not satisfied: RHS $$= \mu (A) \int_X f d \mu$$ which is not the same as LHS. Your computation of RHS is wrong. If you apply Fubini's Theorem correctly to $$\int f(x)I_{A\times X}(x,y)d(\mu \times \mu) (x,y)$$ you will simply get $$\int_A fd\mu=\int_A fd\mu$$!

• Please read the edit, I think I explained it poorly – Mr Martingale Apr 6 at 12:13