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Prove that $EX=E(E(X|Y))$

I know that I should prove it from definition of conditional distribution and conditional expected value, but I don't know how.

I have also looked at theorem("Total expectation") which should be connected with the proof.

If someone would tell me, which properties of expectation I should use to prove it, I would be really glad.

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    Do you want a proof for the discrete case or the general one? Have you studied measure theory? – 2017-01-22
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    In the wiki page of Law of total expectation [here](https://en.wikipedia.org/wiki/Law_of_total_expectation) there is a proof... – 2017-01-22
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    I haven't study measure theory. For discrete one. – 2017-01-22

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\begin{eqnarray*} \mathbb{E}(\mathbb{E}(X|Y)) & = & \int_{-\infty}^{\infty} \mathbb{E}(X|Y = y) f_Y(y) dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X|Y}(x|y) dx \ f_Y(y) dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X|Y}(x|y) f_Y(y) dx dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X,Y}(x,y) dx dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X,Y}(x,y) dy dx \\ & = & \int_{-\infty}^{\infty} x\int_{-\infty}^{\infty}f_{X,Y}(x,y) dy dx \\ & = & \int_{-\infty}^{\infty} xf_{X}(x) dx \\ & = & \mathbb{E}(X) \end{eqnarray*}

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By definition $E(X\mid Y)$ is random variable which satisfies: $$ E(X \cdot 1_G) = E\bigl(E(X\mid Y)\cdot 1_G\bigr), \quad \text{for all} \ G \in \sigma(Y). $$ Simply take $G=\Omega$ such that $1_G=1$ and you get the desired result.