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suppose that ${\mu}_n$ is a sequence of finite measures on measurable space $(X,M)$ where uniformly convergence to $\mu$.
and $\theta$ is a measure on $(X,M)$.

prove that if $\theta$$\bot$$\mu_n$ then $\theta$$\bot$$\mu$.
my work : $\theta$ $\bot$ $\mu_n$ its mean there exist $E_n$ , $F_n$ $\in$ $M$ , where $E_n$ $\cup$ $F_n$ $=$ $X$ and $E_n$ $\cap$ $F_n$ $=$ $\varnothing$ and $E_n$ is $\mu$ null and $F_n$ is $\theta$ null.
now we shall found two appropriate set in $M$ where has above property for $\theta$ and $\mu$.
please help

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    What do you mean by uniform convergence of measures? – 2017-01-15
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    $\mu_n$ is sequence of finite measures where uniformly convergence to $\mu$ – 2017-01-15
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    its mean that there exist a $K$ in $M$ where $\forall$$x$ in $X-K$ , $\mu_n$ convergence uniformly to $\mu$ and $\mu(K)$$=$$0$ – 2017-01-15
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    I still don't get what you mean when you say "$\mu_n$ converges uniformly to $\mu$". I know the concept for functions but not for measures – 2017-01-15
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    Do you mean that the total variation norm of the difference goes to zero? – 2017-01-15
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    its mean $\mu_n$ $\to$ $\mu$ $\forall$$x$ out of a $K$ where $\mu(K)$ $=$ $0$ – 2017-01-15

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