Conditional expectation: Prove $X_1 = X_2 \text{ a.s.}$

Question

(This is a problem in chap 4 of 概率论教程（缪柏其 胡太忠著）. )

Set space $(\Omega, \mathscr{A}, P)$, $\mathscr{C}_1, \mathscr{C}_2 \subset \mathscr{A}$, and $\mathscr{C}_1, \mathscr{C}_2$ are both $\sigma$ algebra, $X\in L_1, X_1=E[X|\mathscr{C}_1], X_2=E[X_1|\mathscr{C}_2]$. If $X=X_2, a.s.$, Prove $X_1=X_2, a.s.$.

I have thought about this for a few days, the point is to show that $X=E[X|\mathscr{C}_1], a.s.$, I have got the following result:

1. $\mathrm{E}\left[ |X_1| \right] =\mathrm{E}\left[ |X_2| \right],\ \mathrm{E}\left[ X_1 \right] =\mathrm{E}\left[ X_2 \right]. $
2. $\mathrm{E}\left[ |X||\mathscr{C} _1 \right] =|\mathrm{E}\left[ X|\mathscr{C} _1 \right] |,\ \mathrm{E}\left[ |X_1||\mathscr{C} _2 \right] =|\mathrm{E}\left[ X_1|\mathscr{C} _2 \right] |$.

I have read about two different answers, but I feel both of them have some bugs, however, it may provide some directions:

1. By the equation condition of Jensen inequality.
2. Try to show $X_1\ge c\,\,\Longleftrightarrow X\ge c,\ \forall c \in \mathbb{R} $.

Answer 1

I have found two methods to solve this problem, these methods come from this question. We can simply modify these methods to solve the problem.

1. Let $h: \mathbb{R}\to\mathbb{R}$ to be a strictly increasing bounded continued function, then $$X_ih\left( X_j \right) \in L_1,\quad i,j=1,2.$$ So $$\mathrm{E}\left[ \left( X_1-X_2 \right) \left( h\left( X_1 \right) -h\left( X_2 \right) \right) \right] \\ =\mathrm{E}\left[ X_1h\left( X_1 \right) \right] -\mathrm{E}\left[ X_2h\left( X_1 \right) \right] -\mathrm{E}\left[ X_1h\left( X_2 \right) \right] +\mathrm{E}\left[ X_2h\left( X_2 \right) \right] ,$$ and since $X_1\in \mathscr{C} _1,\ X_2\in \mathscr{C} _2,$ we have $$ \mathrm{E}\left[ X_2h\left( X_1 \right) \right] =\mathrm{E}\left[ \mathrm{E}\left[ X_2h\left( X_1 \right) |\mathscr{C} _1 \right] \right] \\ =\mathrm{E}\left[ h\left( X_1 \right) \mathrm{E}\left[ X_2|\mathscr{C} _1 \right] \right] \\ =\mathrm{E}\left[ h\left( X_1 \right) \mathrm{E}\left[ X|\mathscr{C} _1 \right] \right] \\ =\mathrm{E}\left[ X_1h\left( X_1 \right) \right], $$ similarly, $\mathrm{E}\left[ X_1h\left( X_2 \right) \right] =\mathrm{E}\left[ X_2h\left( X_2 \right) \right] .$ Thus, $$\mathrm{E}\left[ \left( X_1-X_2 \right) \left( h\left( X_1 \right) -h\left( X_2 \right) \right) \right] =0.$$ So $\left( X_1-X_2 \right) \left( h\left( X_1 \right) -h\left( X_2 \right) \right) =0,\ \mathrm{a}.\mathrm{s}.$ And since $$ \left\{ \left( X_1-X_2 \right) \left( h\left( X_1 \right) -h\left( X_2 \right) \right) =0 \right\} \subset \left\{ X_1-X_2=0 \right\} , $$ we get $X_1=X_2,\ \mathrm{a}.\mathrm{s}.$.

2. $\forall a \in \mathbb{R},$ we have $$ \mathrm{E}\left[ \left( X_1-X_2 \right) I_{\left\{ X_1\leqslant a \right\}} \right] \\ =\mathrm{E}\left[ X_1I_{\left\{ X_1\leqslant a \right\}} \right] -\mathrm{E}\left[ X_2I_{\left\{ X_1\leqslant a \right\}} \right] \\ =\mathrm{E}\left[ X_1I_{\left\{ X_1\leqslant a \right\}} \right] -\mathrm{E}\left[ \mathrm{E}\left[ X_2I_{\left\{ X_1\leqslant a \right\}}|\mathscr{C} _1 \right] \right] \\ =\mathrm{E}\left[ X_1I_{\left\{ X_1\leqslant a \right\}} \right] -\mathrm{E}\left[ X_1I_{\left\{ X_1\leqslant a \right\}} \right] =0. $$ Also, $$ \mathrm{E}\left[ \left( X_1-X_2 \right) I_{\left\{ X_1\leqslant a \right\}} \right] \\ =\mathrm{E}\left[ \left( X_1-X_2 \right) I_{\left\{ X_1\leqslant a,X_2>a \right\}} \right] +\mathrm{E}\left[ \left( X_1-X_2 \right) I_{\left\{ X_1\leqslant a,X_2\leqslant a \right\}} \right] = 0 , $$ since $\mathrm{E}\left[ \left( X_1-X_2 \right) I_{\left\{ X_1\leqslant a,X_2>a \right\}} \right] \leqslant 0$, we have $\mathrm{E}\left[ \left( X_2-X_1 \right) I_{\left\{ X_1\leqslant a,X_2\leqslant a \right\}} \right] \geqslant 0,$ which leads to $\mathrm{P}\left( X_1\leqslant a,X_2>a \right) =0.$ This further leads to $$ \mathrm{P}\left( X_2>X_1 \right) =\mathrm{P}\left( \bigcup_{a\in \mathbb{R}}{\left\{ X_1\leqslant a,X_2>a \right\}} \right) =0. $$ Similarly, $\mathrm{P}\left( X_1>X_2 \right)=0$, thus $X_1=X_2,\ \mathrm{a}.\mathrm{s}.$

Again these methods come from others, you can check the link to learn the original version.