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I recently encountered a version of the Vitali - Hahn - Saks Theorem in the book "Measure Theory" by S. Negrepontis stated as follows:

Let $(X,\mathcal{A})$ be a measurable space and $\{\mu_n\}$ a sequence of finite measures on $(X,\mathcal{A})$ such that $\tau(A)=\lim\limits_{n}\mu_n(A)<\infty$ exists for each $A \in \mathcal {A}$. Then $\tau$ is a measure on $(X,\mathcal{A})$.

I have a proof for the statement above but the author gives a different-and more complex-one so I'm wondering if there is a gap in mine. I present it here.

Proof: It is easily obtained that $\tau(\emptyset)=0$ and $\tau$ is finitely additive. So, it suffices to show that for each descending sequence $\{A_n\}$ such that $\bigcap\limits_{n=1}^{\infty}A_n=\emptyset$ it holds that $\lim\limits_{n}\tau(A_n)=0$. Suppose there exists a sequence $\{A_n\}$ such that $\bigcap\limits_{n=1}^{\infty}A_n=\emptyset$ and $\lim\limits_{n}\tau(A_n)>0$ (which exists since $\tau(A_n)$ is non-increasing). Then, there exists $\epsilon_1>0$ and $N_0$ such that $\lim\limits_{m}\mu_m(A_n)\geq \epsilon_1>0$ for each $n\geq N_0$. Owing to the latter, there exists $\epsilon_2>0$ and $M_0$ such that $\mu_m(A_n)>\epsilon_2>0$ for each $m \geq M_0$, $n \geq N_0$. Then, for the descending sequence $\{B_n\}$ given by $B_n=A_{N_0+n}$ we have $\lim\limits_{n}\mu_{M_0}(B_n)\geq \epsilon_2 >0$ and $\bigcap\limits_{n=1}^{\infty}B_n=\emptyset$ with $\mu_{M_0}$ a finite measure, a contradiction.

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