# A Version of Vitali - Hahn - Saks Theorem.

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.