I'm having a bit of trouble in proving the following exercise in Halmos' "Measure Theory" (Sec. 32, Exercise 5b, adapted for context):
Given a sequence $\{\mu_n\}$ of totally finite measures, write: $$\overline{\mu}_n=\sum_{i=1}^n\mu_i \quad,\quad\mu=\sum_{i=1}^\infty\mu_i$$ Assume $\mu$ is a finite measure. If $\{\phi_n \}$ and $\{\psi_n \}$ are sequences of funcions such that $\forall n$, $\phi_n=\psi_n$, $\mu_n$-almost everywhere, prove that for $\mu$-almost everywhere: $$\limsup_n\phi_n(x)=\limsup_n\psi_n(x)$$ $$\liminf_n\phi_n(x)=\liminf_n\psi_n(x)$$ (Hint: Write $E_n=\{x:\phi_n(x)\neq\psi_n(x)\}$ and apply the result from (5a): If $\{E_n\}$ is a sequence of measurable sets such that $\overline\mu_n(E_n)=0 $, then $\mu(\limsup_nE_n)=0$.)
I can see how the hint would easily lead to the answer, but I'm having trouble in asserting that:
$$\mu_n(E_n)=0\implies\sum_{i=1}^n\mu_i(E_n)=0$$
Am I missing something? If needed, I can edit in my solution for (5a), but the full solution doesn't seem necessary.
Since the assertion is wrong as written, I'm confident that it is a typographical error, and the assumption ought to be that $\phi_n = \psi_n$ holds $\overline{\mu}_n$-almost everywhere. Then the hint works.
To see that as typeset the assertion is wrong, consider a sequence $(x_n)$ of distinct points and let $\mu_n$ be the point mass with weight $2^{-n}$ at $x_n$. Since $\sum 2^{-n} = 1 < +\infty$, $\mu$ is a finite measure. Now let $\phi_n \equiv 1$ for all $n$, and $\psi_n = \chi_{\{x_n\}}$. Then $\phi_n = \psi_n$ holds $\mu_n$-a.e., but
$$\lim_{n\to\infty} \psi_n(x) = 0 < 1 = \lim_{n\to\infty} \phi_n(x)$$
for all $x$.