Consider $n$ hypothesis testing with size $\alpha/n$. Suppose there are $n_0$ of them are true. Let $V$ be the random number of falsely rejected hypothesis of those true ones.
$\textbf{Q:}$ How to show expected number of false rejection $E[V]\leq \sum_{i,H_0 true} Pr(p_i\leq\frac{\alpha}{n})$ where sum is over those true hypothesis, $Pr(-)$ is the probability $p_i$ is the p-value. One can see that $E[V]=\sum_{v\leq n_0}P(V\geq v)$. This is shown in the following document's 4.4.1. https://statweb.stanford.edu/~candes/teaching/stats300c/Lectures/Lecture04.pdf
I don't think this is quite right.
Let $I$ be the set of indices corresponding to the true null hypotheses, so there are $n_0$ elements in $I$.
And suppose $\hat p_i$ is the p-value corresponding to the $i$th null hypothesis $H_{0i}$, $i=1,2,\ldots,n$.
Then, $$V=\sum_{i\in I}I\left(\hat p_i\le \frac{\alpha}{n}\right)$$
Therefore,
$$E\left[V\right]=\sum_{i \in I}E\left[I\left(\hat p_i\le \frac{\alpha}{n}\right)\right]=\sum_{i \in I} P\left(\hat p_i\le \frac{\alpha}{n}\right)\le \sum_{i \in I} \frac{\alpha}{n}=\frac{n_0\alpha}{n}$$
The only inequality used here is $P(\hat p_i\le x)\le x$ for $0\le x\le 1$.