When can we take infinite limits accross an inequality? $\newcommand{\scrF}{\mathscr{F}}$
I'm trying to understand the following statement and it's proof. First some definitions
Definitions: If $\Omega$ is a set and $\mu^\ast$ is an outer measure defined on $\Omega$, then we denote $\scrF_{\mu^\ast}$ to be the collection of all $\mu^\ast$ measurable sets in $\Omega$. Note that this is a field in $\Omega$.
Proposition: Let $\Omega$ be a set and $\mu^\ast$ an outer measure defined on $\Omega$. Let $(E_n)$ be a sequence of disjoint sets in $\scrF_{\mu^\ast}$, then for any $T \subseteq \Omega$ we have that
$$
\mu^\ast \left(T \cap \left(\bigcup_{n=1}^\infty E_n \right)\right) = \sum_{n=1}^\infty \mu^\ast (T \cap E_n).
$$
To prove the statement we argue via an induction on $n$ to arrive at
$$
\mu^\ast \left(T \cap \left(\bigcup_{n=1}^k E_n \right)\right) = \sum_{n=1}^k \mu^\ast (T \cap E_n), (\forall k \in \mathbb{N}).
$$
Getting here is fine to me, but to finish the proof I have an issue. We then note that
$$
T \cap \left(\bigcup_{n=1}^\infty E_n \right) \supseteq T \cap \left(\bigcup_{n=1}^k E_n \right),
$$
and so by monotonicity of the outer measure we have that
$$
\mu^\ast \left(T \cap \left(\bigcup_{n=1}^\infty E_n \right)\right) \geq \mu^\ast \left(T \cap \left(\bigcup_{n=1}^k E_n \right)\right) = \sum_{n=1}^k \mu^\ast(T \cap E_n),
$$
where the equality is from our induction. We next say that letting $k \rightarrow \infty$ on the RHS gives the desired inequality. There's probably some basic Analysis that I'm forgetting, but why do we get to just let $k$ approach infinity in the limit here; i.e what guarantees nothing goes wrong? Thanks in advance for the clarification.
 A: The "basic result" you may be forgetting is that for any convergent sequence $(a_k) \subseteq \mathbb{R}$, we have the following: if there exists a $b \in \mathbb{R}$ such that $a_k \leq b$ for every $k$, then $\lim_{k\to\infty} a_k \leq b$ also. This limit inequality also holds trivially if $b = \infty$, or if any $a_k$ is actually infinite.
In your case, $b$ is the fixed (extended) real number $\mu^*\left(T\cap\left(\bigcup_{n=1}^\infty E_n\right)\right)$, and $(a_n)$ is the sequence of partial sums $a_k = \sum_{n=1}^k \mu^*(T\cap E_n)$. By definition of an infinite series,
$$
\sum_{n=1}^\infty \mu^*(T\cap E_n) = \lim_{k\to\infty} \sum_{n=1}^k \mu^*(T\cap E_n) = \lim_{k\to\infty} a_k.
$$
So because you showed $a_k \leq b$ for all $k$, the above limit theorem applies to give us
$$
\mu^*\left(T\cap\left(\bigcup_{n=1}^\infty E_n\right)\right) = b \geq 
\lim_{k\to\infty} a_k =\sum_{n=1}^\infty \mu^*(T\cap E_n).
$$
By the way, if you were concerned about the inequality going only one way, the other way follows by subadditivity:
$$
\mu^*\left(T\cap\left(\bigcup_{n=1}^\infty E_n\right)\right) = \mu^*\left(\bigcup_{n=1}^\infty (T \cap E_n)\right) \leq \sum_{n=1}^\infty \mu^*(T\cap E_n).
$$
A: Recall that for a non-decreasing sequence of real numbers, we have $\lim a_n = \sup \{a_n\}$, and an analogous statement holds for non-increasing sequences. In particular, $\limsup a_n$ is the limit of a certain non-increasing subsequence of $a_n$, and is thus the supremum of said subsequence. Hence if $x\geq a_n$ for all $n$, then $x\geq \sup \{a_n\}=\limsup a_n$.
