Proof that conditional probabilities sum up to one I've read on Wikipedia that the sum (or integral, for continuous $Y$) of $P(Y=y|X=x)$ over $y$ is always equal to one. I've attempted a proof of this statement for the discrete (sum) case:
Proof: By the Kolmogorov definition of conditional probability and the Law of Total Probability, 
$$\sum_k P(A_k |  B) = \sum_k \frac{P(A_k \cap B)}{P(B)} = \frac{1}{P(B)}\sum_kP(A_k \cap B) = \frac{1}{P(B)}P(B)=1.\ \square$$
Is this a correct proof? How would I proof the above statement for the continuous case, i.e. that $\int_y P(Y=y|X=x)$ always equals one?

Your proof looks great! For the continuous case, your probability is actually a density, so you could write
\begin{align*}
\int_{\mathbb{R}} f(y \mid x) dy = \int_{\mathbb{R}} \frac{f(y,x)}{f(x)} dy = \frac{1}{f(x)}\int_{\mathbb{R}} f(y,x) dy = 1
\end{align*}
since the integral of the joint density over the support of $Y$ is simply the marginal for $X$, so the proof is more or less the same.
