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I need to prove the total expectation theorem using the Lebesgue integration only. I have seen the proof in some forums and some probability books, but they use the usual integration encountered in the calculus based probability books. I want to prove it using the Lebesgue integration. This is the proof that I need to express using the Lebesgue integral

$E[E[X|Y]]$ = $\int ^\infty _{-\infty} E[X|Y=y] \times f_{Y} (y) dy $
$\quad$ $\quad \quad \quad \ $= $\int^\infty _{-\infty} \int^\infty _{-\infty} xf_{X|Y=y}(x) dx f_{Y} (y) dy$
$\quad$ $\quad \quad \quad \ $= $\int^\infty _{-\infty} \int^\infty _{-\infty} x f_{XY}(x,y) dx dy$
$\quad$ $\quad \quad \quad \ $= $\int^\infty _{-\infty} x \int^\infty _{-\infty} f_{XY} (x,y) dy dx$
$\quad$ $\quad \quad \quad \ $= $\int^\infty _{-\infty} xf_{X}(x) dx$
$\quad$ $\quad \quad \quad \ $= $E[X]$

user 2000
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    Very carefully... – Nap D. Lover Jan 02 '18 at 15:45
  • What is the "total expectation theorem"??? – David C. Ullrich Jan 02 '18 at 17:09
  • it is the expectation of the conditional expectation. it basically states that we can find the expectation of a random variable by finding the expectation of the conditional expectation of this random variable by conditioning on another random variable – user 2000 Jan 02 '18 at 17:25
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    In general you cannot assume that $X$ has a conditional density given $Y$ as in the alleged proof in your question. Now the proof for the general case depends on how you define the conditional expectation. But under the usual definition of the conditional expectation, which is that $\tilde{X} = \mathsf{E}[X\mid Y]$ is the $\sigma(Y)$-measurable random variable satisfying $$\forall A\in\sigma(Y),:\quad\mathsf{E}[X\mathbf{1}_A]=\mathsf{E}[\tilde{X}\mathbf{1}_A],$$ You can simply plug $A=\Omega$ to obtain the proof. – Sangchul Lee Jan 02 '18 at 17:31
  • Using the usual definition of the conditional expectation, how can I link it to the conditional distribution to be used in the proof? and for the expectation using the Lebesgue integral, $E[X] = \int X dP$ , how can I use this expectation in the proof, I mean how can I integrate with respect to x and to y, because at the beginning, I am finding the expectation of the conditional expectation with respect to y, so I am integrating with respect to y , then I find the conditional expectation so I am integrating with respect to x . How can I express this using the Lebesgue integral? I – user 2000 Jan 02 '18 at 18:01

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