Semifinite measure induced by a measure Define $\mu_0: \mathcal{M} \longrightarrow [0,\infty]$ by
$$
\mu_0(E) = \sup\{ \mu(F) \mid F \subseteq E, \mu(F) < \infty \},
$$
prove that $\mu_0$ is a measure on $\mathcal{M}$.
My attempt: Obviously $\mu_0(\emptyset)=0$. Then let
$E_i \in \mathcal{M}, \, i \in \mathbb{N}^+$ be pairwisely
disjoint.
$$
\begin{aligned}
\mu_0(\bigsqcup_{i=1}^{\infty}E_i) =&  
\sup \{ \mu(F) \mid F \subseteq \bigsqcup_{i=1}^{\infty}E_i,
\mu(F) < \infty \} \\
%%%%%%%%%%%%%%
%%%%%%%%%%%
=& \sup \{ \mu(\bigsqcup_{i=1}^{\infty}(F \cap E_i)) \mid F \subseteq \bigsqcup_{i=1}^{\infty}E_i,
\mu(F) < \infty \} \\
%%%%%%%%%%%%%%
%%%%%%%%%%%%
=& \sup \{ \sum_{i=1}^{\infty}\mu(F \cap E_i) \mid F \subseteq \bigsqcup_{i=1}^{\infty}E_i,
\mu(F) < \infty \} \\
%%%%%%%%%%%%
%%%%%%%%%%%%%%%
\leq& \sup \{ \sum_{i=1}^{\infty}\mu(F_i) \mid F_i \subseteq E_i,
\mu(F_i) < \infty, \forall i \in \mathbb{N}^+ \}  \\
%%%%%%%%%%%%
%%%%%%%%%%%%%
=& \sum_{i=1}^{\infty} \sup \{\mu(F_i) \mid F_i \subseteq E_i,
\mu(F_i) < \infty \}  \\
%%%%%%%%%%
%%%%%%%%%%%
=& \sum_{i=1}^{\infty}  \mu_0(E_i)  \; .
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
\sum_{i=1}^{\infty}  \mu_0(E_i) =&  
\sum_{i=1}^{\infty} \sup \{\mu(F_i) \mid F_i \subseteq E_i,
\mu(F_i) < \infty \}  \\
%%%%%%%%%%%%%%
%%%%%%%%%%%
=& \sup \{ \sum_{i=1}^{\infty}\mu(F_i) \mid F_i \subseteq E_i,
\mu(F_i) < \infty, \forall i \in \mathbb{N}^+ \} \\
%%%%%%%%%%%%%%
%%%%%%%%%%%%
%%%%%%%%%%%%%%%
=& \sup \{ \mu(\bigsqcup_{i=1}^{\infty}F_i) \mid F_i \subseteq E_i,
\mu(F_i) < \infty, \forall i \in \mathbb{N}^+ \} \;.
\end{aligned}
$$
But we can't get
$$
\begin{aligned}
&  \sup \{ \mu(\bigsqcup_{i=1}^{\infty}F_i) \mid F_i \subseteq E_i,
    \mu(F_i) < \infty, \forall i \in \mathbb{N}^+ \}   
 \\
%%%%%%%%%%%%%%
%%%%%%%%%%%
\leq& \sup \{ \mu(F) \mid F \subseteq \bigsqcup_{i=1}^{\infty}E_i,
\mu(F) < \infty \} \;, 
\end{aligned}
$$
because $\mu(\bigsqcup_{i=1}^{\infty}F_i)$ might be $\infty$.
 A: First of all, verify that if $\mu_0(E_j)=\infty$ for some $j\in \mathbb{N}$ then the equality is trivially satisfied (i.e. $\mu_0\left(\bigsqcup_i E_i\right)=\infty)$. Suppose now that $\mu_0(E_i)<\infty$ for all $i\in \mathbb{N}$, and let $\varepsilon > 0$. By the characterization of the supremum, we can find sets $F_i \subseteq E_i$ with $\mu(F_i) < \infty$ and
$$\mu_0(E_i) - \frac{\varepsilon}{2^i} \leq \mu(F_i) \leq \mu_0(E_i)$$
Summing the above for all $i \in \mathbb{N}$, we have
$$\sum_{i=1}^\infty \mu_0(E_i) - \varepsilon \leq \sup\left\{\mu\left(\bigsqcup_{i=1}^nF_i\right): n\in \mathbb{N}\right\} \leq  \sup \left\{ \mu(F) \mid F \subseteq \bigsqcup_{i=1}^{\infty}E_i,
    \mu(F) < \infty \right\} = \mu_0\left(\bigsqcup_{i=1}^\infty E_i\right)$$
where we used the fact that
$$\sum_{i=1}^\infty \mu(F_i) = \lim_{n\to\infty}\sum_{i=1}^n \mu(F_i) = \lim_{n\to \infty}\mu\left(\bigsqcup_{i=1}^n F_i\right)=\sup\left\{\mu\left(\bigsqcup_{i=1}^nF_i\right):n\in \mathbb{N}\right\}$$
Summarizing, for all $\varepsilon > 0$, we have
$$\sum_{i=1}^\infty \mu_0(E_i) - \varepsilon \leq \mu_0\left(\bigsqcup_{i=1}^\infty E_i\right)$$
which proves the required inequality.
