Does the base measure in Dirichlet Process normalize to one? In the formal definition of Dirichlet process, we know that $\sum_{k=1}^nX(B_k)=1$ by the property of Dirichlet distribution. My question is, is it true that $\sum_{k=1}^nH(B_k)=1$? Why?
 A: The statement $\sum_{k=1}^n H(B_k) = 1$ is true because the base distribution $H$ is a probability distribution and so its sum over all realizations must be equal to one. Since this isn't a very satisfying answer, let's look at the definition of the DP from that Wikipedia page. The formal definition is:
$(X(B_1),...,X(B_n)) \sim Dir((\alpha H(B_1),...,\alpha H(B_n))$
Now, if we look at another definition of the Dirichlet distribution referenced in the $Dir$ above at https://en.wikipedia.org/wiki/Dirichlet_distribution , we see that the function's argument is  a vector of positive values $ \alpha_1,...\alpha_k$. We can sum that up into a scalar concentration parameter $\alpha$ and the vector of probabilities $H$ such that $\alpha\sum_k^K(H_k) = \sum_k^K \alpha_kJ_k$, where $J$ is some other probability distribution which is being modified by those $\alpha$ coefficients Sorry for double dipping on $\alpha$ here; the two wikipedia pages have two slightly different meanings of that symbol (vector versus scalar).
One way to understand the Dirichlet process is the 'stick breaking' construction. I encourage you to read about that here:
https://en.wikipedia.org/wiki/Dirichlet_process#The_stick-breaking_process .
In this representation, if the sum of $H$ exceeds 1, then the DP would be trying to 'break off' parts of the stick which we didn't have to begin with.
In short, it is true because we defined $H$ to be a base probability distribution which is then rescaled using the concentration parameter which can be either a scalar for a symmetric Dirichlet distribution or a vector for a nonsymmetric one.
