Prove $\mu(V)=\mu(E) $, when $V = \cap_{k=1}^{\infty} {V_k} $ Prove  $\mu(V)=\mu(E)$, when $V = \cap_{k=1}^{\infty} {V_k}$
$E \subset  {V_k}$,For each k , $\mu(V_k) < \mu(E) + \epsilon\cdot 2^{-k}$
 A: Since $V=\bigcap_{i=1}^{\infty} V_i\subseteq V_k$ for any positive integer $k$, we have $$\mu(V)\leq \mu(V_k)<\mu(E)+\frac{\varepsilon}{2^k}.$$ Since the strict inequality $$\mu(V)<\mu(E)+\frac{\varepsilon}{2^k}$$ is true for all $k$, its weak version must be true in the limit: $$\mu(V)\leq\mu(E)+\lim_{k\to\infty}\frac{\varepsilon}{2^k}=\mu(E).$$
As for the other direction, $E$ is contained in $V_k$ for all $k$, so it must be contained in their intersection as well: $E\subseteq V$. By the monotonicity of measures, we have $\mu(E)\leq \mu(V)$.
Putting the two weak inequalities derived above together, we conclude that $\mu(V)=\mu(E)$.

In the proof above, I used the following
$\textbf{Lemma}\quad$ Suppose $a\in\mathbb{R}$ and $(b_k)_{k\in\mathbb{Z}_+}$ is a sequence of real numbers such that
(i) $a<b_k$ for all positive integers $k$; and
(ii) $\lim_{k\to\infty}b_k$ exists and is equal to $b$.
Then, $a\leq b$.
$\textit{Proof}\quad$ Suppose (i) and (ii) are true, but $a>b$. I will derive a contradiction. If $a>b$, let $\varepsilon=a-b>0$. Since $(b_k)$ converges to $b$ by (ii), there exists a positive integer $K$ such that for any positive integer $k>K$, the following must be true: $$|\,b_k-b\,|<\varepsilon.$$ But then, $$b_k-b\leq |\,b_k-b\,|<\varepsilon=a-b,$$ that is, $b_k<a$, which contradicts (i). Hence, supposing $a>b$ has led to a contradiction, so that if (i) and (ii) are true, then $a\leq b$ must necessarily be true as well.
A: Since $E \subset V_k$ for all $k$, it is also a subset of $V$. Therefore $\mu(E)\le\mu(V)$ by $\sigma$-subadditivity. Since $V\subset V_j$ for all $j$, we know that $\mu(V) \le \mu(V_j) < \mu(E) + \varepsilon\cdot2^{-j}$ for all $j$. However as $j$ tends to infinity, this leaves us with $\mu(E) \le \mu(V) \le \mu(E)$. Therefore $\mu(E) = \mu(V)$.
