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It is well known that if $\mu, \mu_k$ $(k = 1,2,\ldots)$ are Radon measures on $\mathbb{R}^n$, then the following statements are equivalent:

  1. $\lim_{k \to \infty}\int_{\mathbb{R}^n}\varphi\, d\mu_k = \int_{\mathbb{R}^n}\varphi\, d\mu$ for all $\varphi \in C_c(\mathbb{R}^n)$.
  2. $\lim_{k \to \infty}\mu_k(B) = \mu(B)$ for each bounded Borel set $B \subset \mathbb{R}^n$ with $\mu(\partial B) = 0$.

My question is the following:

Let $\mu, \mu_k$ $(k = 1,2,\ldots)$ be Radon measures on $\mathbb{R}^n$, and let $f_{k}, f: \mathbb{R}^n \to \mathbb{R}$, with $|f_{k}| = 1$ $\mu_k$ a.e., and $|f| = 1$ $\mu$ a.e. (so that $f_kd\mu_k$, $fd\mu$ are signed Radon measures) and suppose we know that

\begin{equation} \lim_{k \to \infty}\int_{\mathbb{R}^n}\varphi f_k \,d\mu_k = \int_{\mathbb{R}^n}\varphi f\, d\mu \quad\text{ for all } \varphi \in C_c(\mathbb{R}^n). \end{equation}

Is it true that

\begin{equation} \lim_{k \to \infty}\int_{B}f_k\,d\mu_k = \int_{B}f\,d\mu \end{equation} for each bounded Borel set $B \subset \mathbb{R}^n$ with $\mu(\partial B) = 0$?

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  • $\begingroup$ What does $|\nu_{k}| = 1$ mean? Is $|\cdot|$ some norm? $\endgroup$ – Michael Greinecker Apr 1 '15 at 10:18
  • $\begingroup$ I've slightly simplified the question and changed notation ($\nu_k$ is now $f_k$) : $|\cdot|$ is now just absolute value. $\endgroup$ – Lost in a Maze Apr 1 '15 at 17:45
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No. Signed measures do not behave that nicely.

Indeed, for each $k$ let $\mu_{k} =\delta_{1-2^{-k}} + \delta_{1+2^{-k}}$, and let $f_{k} = f = \begin{cases} 1 & x \le 0 \\ 0 & x > 0 \end{cases}$.

Then $$ \lim_{k \to \infty} \int_{\mathbb{R}} \varphi f_{k} d \mu_{k} = \varphi(1 - 2^{-k}) - \varphi(1 + 2^{-k}) \xrightarrow{k \to \infty} 0 \qquad \forall \varphi \in C_{c}(\mathbb{R}) $$ So that for $\mu$ being the unique measure defined by $\mu(A) = 0$ for all $A \subset \mathbb{R}$ we have $$ \lim_{k \to \infty} \int_{\mathbb{R}} \varphi f_{k} d \mu_{k} = \int_{\mathbb{R}} \varphi f d \mu \qquad \forall \varphi \in C_{c}(\mathbb{R}) $$

Clearly $\mu(\partial B) = 0$ for $B = [0,1]$. But, $$ \lim_{k \to \infty} \int_{B} f_{k} d \mu_{k} = 1 \neq 0 = \int_{B} f d \mu. $$

Notably, the total variation measure $|f_{k} \mu_{k}|$ do not behave nicely with the total variation measure of $|f \mu|$ which is why the conclusion fails.

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