We prove the discrete case; the continuous case follows analogously using integrals.

Discrete Case (Partition Formula): Let $\{B_1, \ldots, B_n\}$ be a partition with $P(B_i) > 0$ for all $i$.

By definition of conditional expectation: $E[X|B_i] = \sum_x x \cdot P(X = x | B_i)$

Therefore: $\sum_{i=1}^n E[X|B_i] P(B_i) = \sum_{i=1}^n \left[\sum_x x \cdot P(X = x | B_i)\right] P(B_i)$

Rearranging the sums: $= \sum_x x \sum_{i=1}^n P(X = x | B_i) P(B_i)$

By the law of total probability: $\sum_{i=1}^n P(X = x | B_i) P(B_i) = P(X = x)$

Substituting: $= \sum_x x \cdot P(X = x) = E[X]$ □

General Case ($E[E[X|Y]]$): When conditioning on a random variable $Y$ rather than a partition, we use the fact that $E[X|Y]$ is itself a random variable (a function of $Y$).

For discrete $Y$ taking values $y_1, y_2, \ldots$: $E[E[X|Y]] = \sum_j E[X|Y = y_j] \cdot P(Y = y_j)$

This is precisely the partition formula with $B_j = \{Y = y_j\}$.

For continuous $Y$ with density $f_Y$: $E[E[X|Y]] = \int_{-\infty}^{\infty} E[X|Y = y] \cdot f_Y(y) \, dy$

By the definition of conditional expectation: $E[X|Y = y] = \int_{-\infty}^{\infty} x \cdot f_{X|Y}(x|y) \, dx$

Therefore: $E[E[X|Y]] = \int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} x \cdot f_{X|Y}(x|y) \, dx\right] f_Y(y) \, dy$

Interchanging the order of integration (by Fubini's theorem): $= \int_{-\infty}^{\infty} x \left[\int_{-\infty}^{\infty} f_{X|Y}(x|y) f_Y(y) \, dy\right] dx$

Since $f_{X|Y}(x|y) f_Y(y) = f_{X,Y}(x,y)$ (joint density): $= \int_{-\infty}^{\infty} x \left[\int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy\right] dx$

The inner integral gives the marginal density $f_X(x)$: $= \int_{-\infty}^{\infty} x \cdot f_X(x) \, dx = E[X]$ □