Is the limsup of a sequence of measures also a measure?

Given a sequence $(\mu_{n})_{n\in\mathbb{N}}$ of $\sigma$-finite measures on the measurable space $(Ω,\Sigma)$, is the $\limsup_{n}\mu_{n}$ also a $\sigma$-finite measure?

Clearly, $\limsup_{n}\mu_{n}(A)\geq 0$ for all $A\in \Sigma$, and that $\limsup_{n}\mu_{n}(\varnothing)=0$. But I am not sure if $\sigma$-additivity holds.

I have read about the Vitali-Hahn-Saks theorem which shows that the limit of a sequence of measures is also a measure. Can we adapt the proof of Vitali-Hahn-Saks' theorem or is there a counterexample?

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Consider a space with two elements $\{ x, y \}$. Then

$$\Sigma= \{ \emptyset , \{x \} , \{ y \} , \{ x,y \} \}\,.$$

Define

$$\mu_{2n}= \delta_x \,;\, \mu_{2n+1}=\delta_y \,.$$

Then

$$\limsup \mu_n ( \{ x \} ) =1, \limsup \mu_n ( \{ y \} ) =1,$$ $$\limsup \mu_n ( \{ x,y \} ) =1$$

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