The conditional expectation of a positive random variable is positive

Question

How can the following property of conditional expectation can be proven: $X \ge 0 \implies E[X|G] \ge 0$?

Answer 1

$$ E[X | G] = \int_{-\infty}^{\infty} xf_{x|g}(x|g)dx $$

$X \geqslant 0$, and $ 0 \leqslant f_{x|g}(x|g) \leqslant 1$ because $f_{x|g}(x|g)$ it is a probability distribution function. Thus, $E[X | G] \geqslant 0$ because it is a sum of products of non-negative numbers.

The discrete case is similar:

$$ E[X | G] = \sum_x xf(x|G) $$

Similarly, this is a sum of products of non-negative numbers, so $E[X|G] \geqslant 0$.

Answer 2

Hint: If $\Bbb E[Y\Bbb 1_E]\ge0$ for every measurable set $E$ then $Y\ge0$ almost surely.