Combining two almost universal hash functions We say that a set of functions $S$ is $\epsilon$-almost-universal if for any function $f \in S$, it holds that $\Pr[f(m) = f(m')] &lt; \epsilon$ for $m \not = m'$.

Assume that we have the following two sets of hash functions:

$F = \{f:M \rightarrow X\}$, which is $\epsilon_1$-almost-universal.
$G = \{g:X \rightarrow Y \}$, which is $\epsilon_2$-almost-universal.  

Then, there exists a third set of hash functions $H$ for that it holds:
$H=\{h:M \rightarrow Y\}$ where $h(M)=g(f(M))$ and $H$ is $(\epsilon_1 + \epsilon_2)$-almost-universal.

Now my question is: Why is this the case? Why is $H$ $(\epsilon_1 + \epsilon_2)$-almost-universal?
Under what circumstances is $g(f(m)) = g(f(m'))$ for any $m \ne m'$?  With what probability do these circumstances happen?

(Hint: Write $x = f(m)$ and $x' = f(m')$ and consider the probability of $g(x) = g(x')$.)