Proof of continuity from above in measure theory I started to read about the measure theory in order to refresh my knowledge. When I read the following theorem

Theorem: Given a measure space $(X,\Sigma,\mu)$, if $(A_n)_{n\geq 1}$ is a decreasing sequence of sets in $\Sigma$ with $\mu(A_N)<\infty$ for some $N\geq 1$, then
\begin{equation*}
\mu\left ( \bigcap_{n=1}^{\infty} A_n \right )=\lim_{n\to \infty}\mu(A_n).
\end{equation*}

there was no proof of it. In the mathematical literature I could find, they assume that "$\mu(A_1)<\infty$" instead of "$\mu(A_N)<\infty$ for some $N\geq 1$". I mimicked their proof for the general case and filled the missing details. Please, let me know what you think

Proof: Define $B_n=A_1\setminus A_{N+n-1}$ for $n\geq 1$. Since $A_n\subseteq A_N$ for all $n\geq N$, we have that $\mu(A_n)\leq \mu(A_N)<\infty$, so we get
\begin{equation*}
\mu(A_N\setminus A_n)=\mu(A_N)-\mu(A_n).
\end{equation*}
Since $B_n\in \Sigma$ and $B_n\subseteq B_{n+1}$ for all $n\geq 1$, we get by the continuity from below that
\begin{equation*}
\mu\left (\bigcup_{n=1}^{\infty} B_n\right )=\lim_{n\to \infty}\mu(B_n)=\mu(A_N)-\lim_{n\to \infty}\mu(A_{N+n-1})=\mu(A_N)-\lim_{n\to \infty}\mu(A_n).
\end{equation*}
Here, I think the last step is valid, because $(A_n)_{n\geq 1}$ is decreasing, so does $(\mu(A_n))_{n\geq 1}$, which makes it to have a limit, because it is bounded below by $0$. Next, since $\bigcap_{n=1}^{\infty}A_n\subseteq A_N$, we have $\mu\left (\bigcap_{n=1}^{\infty}A_n\right )<\infty$, and so
\begin{equation*}
\mu\left (\bigcup_{n=1}^{\infty}B_n\right )=\mu\left (A_N\setminus \bigcap_{n=1}^{\infty}A_n\right)=\mu(A_N)-\mu\left (\bigcap_{n=1}^{\infty}A_n\right),
\end{equation*}
where we used De Morgan's law in the first step and that $\bigcap_{n=1}^{\infty}A_{N+n-1}=\bigcap_{n=N}^{\infty}A_n=\bigcap_{n=1}^{\infty}A_n$. Combining this result with the expression above, the conclusion follows since $\mu(A_N)$ is a finite value. \QED

 A: I think the final step in your process makes sense.
As far as I am concerned, I also once thought about the proof of this theorem, and here is what I did:
Since $A_k$ is decreasing for $k\in[1,\infty)$, then we let $B_{k+1}=A_{k+1}-A_k$, thus $B_k$ is also measurable.
If we denote the lower bound of $A_k$ is ${A}$, then
$$\mu(A) = \lim_{k\to\infty}\mu(B_k) $$
And we also have $A_1=B_1$, $$\mu(A_N)=\bigcup_{n=1}^{N} B_k$$
$$\mu(A)=\bigcup_{n=1}^{\infty} B_k$$
where $B_k$ are disjoint.
We assume that $\mu(A_N)<\infty$ for some $N>1$ then
$$\mu(A_k)=\mu(A_{k-1})-\mu(A_{k}\cap A_{k-1}^c)=\mu(A_N)-\sum_{s=N}^{k}\mu(A_{s}\cap A_{s-1}^c)$$.
For $k>N$, $0\leq \mu(A_k)\leq \mu(A_N)<\infty$, then the limitation of $\mu(A_k)$ exists, thus we yield
$$\lim_{k\to \infty}\mu(A_k)=\mu(A_N)+\lim_{k\to\infty}(\mu (A_N-A_k))$$.
Since countable additivity $\sum_{k=N}^{\infty}\mu(A_k)=\mu(\bigcup_{k=N}^{\infty} A_k)$ holds, thus we yield
$$\lim_{k\to\infty}(\mu (A_N-A_k))=\mu(\bigcup_{k=N}^{\infty}(A_N\cap A_k^c)=\mu(A_N\cap(\bigcup_{k=N}^{\infty}A_k^c)).$$
We use De Morgen Equation:
$$\mu(A_N\cap(\bigcup_{k=N}^{\infty}A_k^c)=\mu(A_N)-\mu(\bigcup_{k=N}^{\infty}A_k^c)=\mu(A_N)-\mu((\bigcap_{k=N}^{\infty}A_k)^c)$$,
$$\mu(A_N)-\mu((\bigcap_{k=N}^{\infty}A_k)^c)=\mu(A_N)+\mu((\bigcap_{k=N}^{\infty}A_k))\geq \mu((\bigcap_{k=N}^{\infty}A_k)).$$
And $$\mu(A_N\cap(\bigcup_{k=N}^{\infty}A_k^c)=\mu(A_N)-\mu(\bigcup_{k=N}^{\infty}A_k^c)\leq\mu(A_N\cap(\bigcup_{k=N}^{\infty}A_k^c)=\mu(A_N)-\mu(\bigcup_{k=1}^{\infty}A_k^c),$$
Since $A_k$ decreasing, for $k<N$, $\cap_{k=1}^{\infty}A_k\leq\cap_{k=N}^{\infty}A_k$,
then we can say
$$\begin{align*}\lim_{k\to\infty}\mu^*\left(A_k\right)\leqslant\mu^*\left(\bigcap_{k=1}^\infty A_k\right)\end{align*}.$$
And the feasibility of the above subtraction operation is determined by $\mu(A_N)<\infty$, and meanwhile according to the set operation:
$$\begin{aligned}\mu^*\left(\bigcap_{k=1}^\infty A_k\right)&\leqslant\mu^*\left(A_k\right),\forall k.\\ \\ \mu^*\left(\bigcap_{k=1}^\infty A_k\right)&\leqslant\lim_{k\to\infty}\mu^*\left(A_k\right)\text{.}\end{aligned}$$
Thus $$\begin{equation*}
\mu\left ( \bigcap_{n=1}^{\infty} A_n \right )=\lim_{n\to \infty}\mu(A_n).
\end{equation*}$$
