Let $X$ be a metric space and $(X,\mathfrak{A},\mu)$ a finite measure space with $\mathcal{B}_X\subset\mathfrak{A}$. Let $\mu$ be regular on $A,B\in\mathfrak{A}$.

Show that $\mu$ is regular on $A\cap B$

Since $X$ is a metric space inner and outer regularity are defined as follows:

  1. $A\in\mathfrak{A}$ is outer $\mu$-regular: $\mu (A) =\inf\{\mu (U) : U\in\mathfrak{A}\mbox{ is open and }A\subset U \}$
  2. $A\in\mathfrak{A}$ is inner $\mu$-regular: $\mu (A)=\sup\{\mu (C) : C\in\mathfrak{A}\mbox{ is compact and }C\subset A\} $

So I would have show that $A\cap B$ is both inner and outer regular.
Let's first attempt inner regularity:

Since $A$ is inner regular, for every $\varepsilon >0$, exists a compact $C_A\in\mathfrak{A}$ s.t $\mu (A) < \mu (C_A)+\varepsilon$. (and analogously for $B$).

We want to establish that for every $\varepsilon >0$ exists a compact $K\subset A\cap B$ s.t $\mu (A\cap B)< \mu (K)+\varepsilon$.

Since $A\cap B\subset A$, therefore $(A\cap B)\setminus C_A\subset A\setminus C_A$ we do have by monotonicity of $\mu$ that $\mu ((A\cap B)\setminus C_A)<\varepsilon$. But we need a compact $K\subset A\cap B$. I'm confused about how to find such a $K$. Hints appreciated.


$C_A \cap C_B$ is a compact subset of $A \cap B$ and since we are in a finite measure space,

$$\mu(A\cap B) = \mu A + \mu B - \mu (A\cup B) < \mu C_A + \mu C_B + \epsilon - \mu(A\cup B) \le \mu C_A + \mu C_B + \epsilon - \mu(C_A\cup C_B) = \mu(C_A \cap C_B) + \epsilon$$

For outer regularity do the same thing but take $U_A \cup U_B$.


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