Conditional expectation w.r.t Lebesgue measure

Question

Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})=((0,1)^{2},\mathcal{B}((0,1)^{2}),\lambda_{2})$, where $\lambda_{2}$ is the Lebesgue measure in $\Omega=(0,1)^{2}$. Then, for $\mathcal{G}=\mathcal{B}((0,1))\times(0,1)$, I want to prove $$\mathbb{E}(X|\mathcal{G})(x,y)=\int_{0}^{1}X(x,t)d\lambda_{1}(t)$$ So far, I have $$\int_{A}\mathbb{E}(X|\mathcal{G})(x,y)d\lambda_{2}=\int_{A}X(x,y)d\lambda_{2}\qquad\forall A\in\mathcal{B}((0,1))\times(0,1)$$

My idea is to let $(\Omega,\mathcal{F},\mathbb{P})=((0,1),\mathcal{B}(0,1),\lambda_{0})\otimes((0,1),\mathcal{B}(0,1),\lambda_{1})$, for $\lambda_{0}$, $\lambda_{1}$ positive and finite measures. Then use Fubini's theorem to get $$\begin{aligned}\int_{\mathcal{B}(0,1)\times(0,1)}\mathbb{E}(X|\mathcal{G})(x,y)d\lambda_{2}&=\int_{\mathcal{B}(0,1)\times(0,1)}X(x,y)d\lambda_{2}\\&=\int_{\mathcal{B}(0,1)}\int_{0}^{1}X(x,t)d\lambda_{1}(t)\qquad\forall y\in(0,1)\end{aligned}$$

But I'm not sure this is correct and I am still left with integrals that I don't want.

Answer 1

Although this is a conditional expectation problem, most of the proof technicalities resemble that of deriving the Fubini 's theorem. It might be helpful to review the related materials.

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For simplicity, write $\mathscr{B}((0, 1))$ as $\mathscr{B}$ and $\mathscr{B}((0, 1)^2)$ by $\mathscr{B^2}$.

You must assume $X$ is an integrable random variable defined on $(\Omega, \mathscr{F}, P)$ so that $X$ is measurable $\mathscr{B}^2$ and $E[|X|] < \infty$. It can be easily verified that that $\mathscr{G}$ is a sub-$\sigma$-field of $\mathscr{B}^2$. Therefore in order to show the statement, we need to check that $\int_0^1 X(x, t)dt$ ($dt$ is a conventional shorthand for $d\lambda_1(t)$) is a version of $E(X|\mathscr{G})$, that is, to verifty that

1. $\int_0^1 X(x, t) dt$ is measurable $\mathscr{G}$.

2. For any $A \in \mathscr{G}$, it holds that $$\int_A X dP = \int_A\left[\int_0^1 X(x, t) dt\right] dP.$$

To show 1, let's prove the following lemma first:

Lemma $1$: If $f(x)$ is measurable $\mathscr{B}$, then when $f$ is viewed as a function defined on $(0, 1)^2$, it is measurable $\mathscr{G}$.

Proof of the Lemma $1$: For any $z \in \mathbb{R}$, we have $$\{(x, y) \in (0, 1)^2: f(x) \leq z\} = \{x \in (0, 1): f(x) \leq z\} \times (0, 1) \in \mathscr{B} \times (0, 1) = \mathscr{G}.$$

Therefore if we can show that $\int_0^1 X(x, t) dt$ is measurable $\mathscr{B}$, then property $1$ holds. We need prepare another lemma:

Lemma $2$: For every $E \in \mathscr{B}^2$, as a function of $x$, $\lambda_1[y: (x, y) \in E]$ is measurable $\mathscr{B}$.

Proof of the Lemma $2$: If $\mathscr{L}$ is the class of sets $E$ in $\mathscr{B}^2$ such that $\lambda_1[y: (x, y) \in E]$ is measurable $\mathscr{B}$, it is not hard to show that $\mathscr{L}$ is a $\lambda$ system. Since the function is $I_A(x)\lambda_1(B)$ for $E = A \times B$, $\mathscr{L}$ contains the $\pi$-system consisting of measurable rectangles. Hence $\mathscr{L}$ coincides with $\mathscr{B}^2$ by the $\pi$-$\lambda$ theorem.

Denote $F(x) = \int_0^1 X(x, t) dt$. First consider nonnegative $X$. If $X = I_E$, since $\int_0^1 X dt = \lambda_1[y: (x, y) \in E]$, by Lemma $2$, $F$ is measurable $\mathscr{B}$. Because of linearity, $F$ is a linear combination of functions measurable $\mathscr{B}$ and hence is itself measurable $\mathscr{B}$. The general nonnegative $X$ is the monotone limit of nonnegative simple functions, since measurability is preserved under limit operations, $F$ is again measurable $\mathscr{B}$. For $X$ that is not necessarily nonnegative, write $X = X^+ - X^-$, $F$ is easily seen again measurable $\mathscr{B}$ by the linearity of integration. Therefore property $1$ is verified.

Property $2$ is much less technical to verify as follows: for the general element $A = B \times (0, 1) \in \mathscr{G}$, we have \begin{align} & \int_A\left[\int_0^1 X(x, t) dt\right] dP \\ = & \int_{B \times (0, 1)} F(x) dxdy \\ = & \int_B \left[\int_0^1 dy\right] F(x) dx \\ = & \int_B F(x) dx \\ = & \int_B \left[\int_0^1 X(x, t) dt\right] dx \\ = & \int_{B \times (0, 1)} X(x, t) dxdt \\ = & \int_A X dP. \end{align}