Rudin, Riesz Representation theorem.

277 Views Asked by At

In the Reisz representation theorem, page 40 real and complex analysis, there's a preliminary that I don't understand.

After defining for each open $V$ the measure $\mu$ is defined as

$$ \mu(V) = \sup \left\{ \Lambda f : f \prec V \right\} $$

from such definition we deduce both monotonicity and for each open $E$ we have

$$ \mu(E) = \inf \left\{ \mu(V) : E \subset V, \;V \; \text{open} \right\}. $$

After this it states

Let $M_F$ the class of all $E \subset X$ which satisfy two conditions : 1. $\mu(E) < \infty$; 2. $\mu(E) = \sup \left\{ \mu(K) : K \subset E, K \; \text{compact} \right\}$

What I don't understand is how can be stated that $M_F$ is not empty.

1

There are 1 best solutions below

0
On

It is clear that $\emptyset\in M_F$, since the only $f$ such that $f\prec\emptyset$ is $f=0$ and so $\mu(\emptyset)=0$. So $M_F$ is nonempty.