Justifying simplifying assumption about message distributions for perfectly secret encryption?
My initial crack at a solution is the following, please let me know what you think

Fix an encryption scheme (Gen, Enc, Dec) satisfying definiton 2.1 for all distributions over
$\mathcal{M}$ that assign non-zero probability to each $m\in
\mathcal{M}$. Fix a distribution $p_{\mathcal{M}}$ over $\mathcal{M}$
for which $\exists m \in \mathcal{M}$ such that $p_{\mathcal{M}}(m) = 0$. Let $N$ denote the set of all such $m$. To avoid triviality, assume $\exists$ at least 2 messages $m_0, m_1 \in \mathcal{M}$ such that $p_{\mathcal{M}}(m_0) &gt; 0$ and $p_{\mathcal{M}}(m_1) &gt; 0$. 

Now, we let $p^* = p_{\mathcal{M}}(m_0)$, and create an "altered"
distribution as follows. Let $p'_{\mathcal{M}}(m) =
p_{\mathcal{M}}(m)$ for all $m \in \mathcal{M} \setminus (N \cup
\{m_0\})$, and let $p'_{\mathcal{M}}(m) = \frac{p(m_0)}{| N \; \cup \;
  \{m_0\} |}$ (that is, the altered distribution is the same as the
original, except it distributes the probability $p_{\mathcal{M}}(m_0)$
evenly amongst $m_0$ and all the messages in $N$). For notational
convenience, let $N' :=  N \; \cup \;
  \{m_0\} $.

Now,
$p'_{\mathcal{M}}$ is a distribution in which all messages have
non-zero probability, so (Gen, Enc, Dec) is perfectly secret with
respect to this distribution. That is, Gen and Enc define a distribution
over the set of ciphertexts $\mathcal{C}$ such that:

$$p'(M = m \mid C = c) = \frac{p'(M=m, C = c)}{p'(C = c)} = p'(M = m)$$

for all $m \in \mathcal{M}, c \in \mathcal{C}$. (Here $p'$ denotes the
joint probability over messages and ciphertexts determined by the
distribution $p'_{\mathcal{M}}$ over $\mathcal{M}$, the distribution
over keys generated by Gen, and the randomness of Enc). Then, I claim, (Gen,
Enc, Dec) must be perfectly secret with respect to the original
distribution, $p_{\mathcal{M}}$ over $\mathcal{M}$. 

Claim 1: For all $m \not \in  N'$, and for all $c \in \mathcal{C}$:
 $$p'(M=m, C=c) = p(M=m,
  C=c)$$
(where $p'$ is as before and $p$ denotes the
joint probability over messages and ciphertexts determined by the
distribution $p_{\mathcal{M}}$ over $\mathcal{M}$, the distribution
over keys generated by Gen, and the randomness of Enc). The reason
this holds is that Gen and Enc act independently of the
  distribution over $\mathcal{M}$. Since $p'_{\mathcal{M}}(m) =
  p_{\mathcal{M}}(m) \forall m \not \in  N'$, the claim follows.

Claim 2: For all $c \in \mathcal{C}$:
$$p(M=m_0, C=c) = \sum_{m \in N'} p'(M=m, C=c) $$
First, we note that for any $c\in \mathcal{C}$ and for any $m \in
\mathcal{M} \setminus N'$, $p(C=c \mid M= m) = p'(C = c \mid M = m)$,
since the randomness of Gen and Enc are completely independent of the
distribution over $\mathcal{M}$, and so the message-conditional
probability of any particular ciphertext is equal. Then, fix $c\in \mathcal{C}$
$$
\begin{align*}
p(M=m_0, C = c) &amp;= p(C=c \mid M=m_0)p(M=m_0) \\
&amp;= p'(C = c \mid M = m_0)p(M = m_0), \text{ by the above discussion} \\
&amp;= p'(C=c)p(M= m_0), \text{ since (Gen, Enc, Dec) is perfectly secret
  for } p'_{\mathcal{M}} \\
&amp;= p'(C=c) \sum_{m \in N'} p'(M = m) \\
&amp;= p'(C=c) \sum_{m \in N'} p'(M = m \mid C = c) \\
&amp;= \sum_{m \in N'} p'(M=m, C=c)
\end{align*}
$$

Thus, since 
$$p(C = c) = \sum_{m \in \mathcal{M}} p(M=m, C=c) = \sum_{m
  \in \mathcal{M}} p'(M=m, C=c) = p'(C=c)$$
 we see that the
distribution over ciphertexts remains the same with respect to both distributions
over $\mathcal{M}$. Combined with Claim 1, we see that 

$$
\begin{align}
p(M =m \mid C=c) &amp;= \frac{p(M=m, C=c)}{p(C=c)} \\
&amp;= \frac{p'(M=m, C=c)}{p'(C=c)} \\
&amp;= p'(M=m \mid C=c) \\
&amp;= p'(M=m) \\
&amp;= p(M=m)
\end{align}
$$

for all $m \in \mathcal{M} \setminus N'$ and all $c \in
\mathcal{C}$. For $m \in N' \setminus \{m_0\}$, 
$$p(M= m \mid C = c) =
0 = p(M=m)$$
. 

What remains to be seen is that $p(M=m_0 \mid C =
c) = p(M =m_0)$. This is the case, since for all $c \in \mathcal{C}$
$$
\begin{align}
p(M=m_0 \mid C = c) &amp;= \frac{p(M=m_0, C=c)}{p(C=c)} \\
&amp;= \sum_{m \in N'} \frac{p'(M=m, C=c)}{p'(C=c)}, \text{ by Claim 2} \\
&amp;= \sum_{m \in N'} p'(M=m \mid C=c) \\ 
&amp;= \sum_{m \in N'} p'(M=m), \text{ since (Gen, Enc, Dec) is
  perfectly secret w.r.t }p'_{\mathcal{M}}\\ 
&amp;= p(M=m_0), \text{ by definition of }p'_{\mathcal{M}}
\end{align}
$$
And so the claim is proven.