Show that $|μ|(C)=\sup_{\substack{f\in C_b\\|f|\le1\\\left.f\right|_{C^c}=0}}\mu f$ for all open $C$

Question

Let $E$ be a metric space and $\mu$ be a finite signed measure on $\mathcal B(E)$.

I would like to show that$^1$ $$|\mu|(C)=\sup_{\substack{f\in C_b(E)\\|f|\le1\\\left.f\right|_{E\setminus C}=0}}\mu f\;\;\;\text{for all open }C\subseteq E\tag1.$$ Moreover, I would like to know whether there is a similar identity for $|\mu|(A)$ and closed $A\subseteq E$.

Note that "$\le$" in $(1)$ is clearly trivial. The other inequality should follow from the following two results:

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Lemma 1: If $\nu$ is a finite signed measure on $\mathcal B(E)$, $B\in\mathcal B(E)$ and $\varepsilon>0$, then $$|\nu|(C\setminus A)<\varepsilon\tag2$$ for some closed $A\subseteq E$ and open $C\subseteq E$ with $A\subseteq B\subseteq C$.

Lemma 2: If $\emptyset\subset A\subseteq E$ and $C\subset E$ with $\overline A\subseteq C^\circ$, then there is a $f\in C_b(E)$ with $0\le1 f\le 1$ and \begin{align}\left.f\right|_A&=1\tag{3a}\\\left.f\right|_{E\setminus C}&=0.\tag{3b}\end{align}

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Now let $C\subseteq E$ be open and $\varepsilon>0$. Let $E^{\pm}$ be a Hahn decomposition of $E$ wrt $\mu$ and $C^\pm:=C\cap E^\pm$. By Lemma 1, applied to both measures in the Jordan decomposition $\mu^\pm$ of $\mu$ individually, $$\pm\mu(C^\pm\setminus A^\pm)=\mu^\pm(C\setminus A)<\varepsilon\tag4$$ for some closed $A^\pm\subseteq E$ with $A^\pm\subseteq C^\pm\subseteq C$. (We may need to exclude the cases $A^\pm=\emptyset$ and $C^\pm=E$.)

Now, by Lemma 2, there is a $f^\pm\in C_b(E)$ with $0\le f^\pm\le 1$ and \begin{align}\left.f^\pm\right|_{A^\pm}&=1\tag{5a}\\\left.f^\pm\right|_{E\setminus C^\pm}&=0.\tag{5b}\end{align} Let $$g:=f^+-f^-.$$ Note that $g\in C_b(E)$ with $|g|\le1$ and \begin{align}\left.g\right|_{A^\pm}&=\pm1\tag{6a}\\\left.g\right|_{E\setminus C}&=0.\tag{6b}\end{align}

Question 1: We are done if we can show that $\mu g>|\mu|(C)-2\varepsilon$. However, for some reason, I'm not able to obtain this inequality. How do we need to split the integral $$\mu g=\int_Cg\:{\rm d}\mu\tag7$$ in order to obtain this result?

Question 2: Can we find a similar identity for $|\mu|(A)$ and closed $A\subseteq E$?

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$^1$ $C_b(E)$ denotes the space of bounded continuous functions and $\mu f:=\int f\:{\rm d}\mu$.

Answer 1

Consider a Hahn decomposition $E = E^{+}\sqcup E^{-}$. For any set, denote $C^{\pm} = C \cap E^{\pm}$. Now, we have $|\mu|(C)= \mu(C^{+}) - \mu(C^{-})$. Now take $A^{\pm} \subset C^{\pm}$, , such that $|\mu|(C^{\pm}\backslash A^{\pm}) < \epsilon$ ( we assume some regularity here). Consider a continuous function $f^{\pm} $ defined on $E$ with values in $[0,1]$, and taking value $1$ on $A^{\pm}$, and $0$ on $A^{\mp}$. Now consider $ g= f^+- f^-$. Multiply it with a function with values in $[0,1]$, taking value $0$ on $E\backslash C$, and $1$ on $A^+\cup A^-$. So may assume from the beginning that $g$ is zero outside $C$. Now, we can see that the integral $\mu(g)$ is close to $|\mu|(C)$.

$\bf{Added:}$ We have $$||\mu|(C)- \int_C g d\mu |= |\int_{C^+\backslash A} (1-g) d \mu + \int_{C^{-} \backslash B} (-1 - g) d\mu| \le 2 \epsilon + 2 \epsilon = 4\epsilon$$

Note: we can do this also for complex measures, or say for measures with values in some finite dimensional Hilbert space $E$. We then have to consider $\mu(f)$ for $f$ continuous with values in $E$ and $\|f\|\le 1$. Instead of considering the Hahn decomposition, we consider the vector density $\rho$ (Radon-Nikodym) of $\mu$ with respect to $\|\mu\|$, which is a measurable $E$-valued function of constant absolute value $1$. Now we have to piece-wise interpolate this $\rho$, like above, but with more values.

$\bf{Added:}$

Some calculations about limits of measures Consider a net of measures $\mu_t$ converging weakly to $\mu$. Then we have $$\lim\inf_t |\mu_t|(U) \ge |\mu|(U)$$ for every $U$ open.

Assume that we have moreover $$|\mu_t|(E)|\to |\mu|(E)$$

Consider $A = E\backslash U$ a closed set.

We have $$\lim\sup_t |\mu_t|(A) = \lim \sup (|\mu_t|(E) - |\mu|(U)) =|\mu|(E) - \lim \sup |\mu_t|(U) \le |\mu|(E) - |\mu|(U)=|\mu|(A)$$

Conversely, if we have the inequality for the closed set $E$, since $E$ is also open, we must have $|\mu_t|(E)\to |\mu|(E)$.

However, we may have a strict inequality for some sequences.

Consider $E=[-1,1]$, $\phi\colon \mathbb{R} \to \mathbb{R}$ support in $[1-,1]$ negative on $(-\infty,0]$, positive on $[0\infty)$ and with $\int_{-1}^1 \phi(x) d x = 1$. Then the sequence of measures $\mu_n(x) = n \phi_n(nx) dx$ converge to the delta measure $\delta_0$, but $|\mu_n| \to C\cdot \delta_0$, with $C = \int_{-1}^1|\phi(x)|dx> 1$