If $\mu_n \rightarrow \mu$, then $\int f d\mu_n$ does not converge to $\int f d\mu$

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Suppose that $\mu_n$ is a sequence of measures converging setwise to a measure $\mu$.

I want to show that 1) if $\mu_n$ is a decreasing sequence of complete measures then $\mu$ is not necessarily complete, 2) if $\mu_n$ is a decreasing sequence of measures, then $\int f d\mu_n$ does not necessarily converge to $\int f d\mu$, where $f$ is a nonnegative measurable function, and 3) in general, $\lim_{n\rightarrow \infty} \int f d\mu_n$ does not exist (but I’m not sure if this is correct).

For 1) here is my counterexample. Let $\mu_n$ be the complete Lebesgue-Stieltjes measure on $\mathbb R$ associated to the nondecreasing function $F(x)=x/n$.Then $\mu_n$ decreases to the zero measure $\mu=0$ . In particular, $\mu[0,1]=0$ but the interval $[0,1]$ contains non measurable sets, showing that $\mu$ is not complete. Is this right? Any easier counterexample?

For 2) take $\mu_n$ as in the case 1, and let $f$ be the function $\chi_{\mathbb R}$. Then $\int d\mu_n=\infty$ but $\int d\mu=0$. Is this counterexample correct? Is there any easier counterexample?

For 3) I couldn’t find any counterexample. Any suggestion?

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For 3) let $\mu_n(A)=\int_A \frac 1 {nx} I_{(1,n+1)} (x)\, dx$ for $n$ even, $\mu_n(A)=\int_A \frac 2 {nx} I_{(1,n+1)} (x)\, dx$ for $n$ odd. Then $\mu_n(A) \to 0$ for each $A$ but $\int |x|d\mu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .