# Real-valued continuous functions with compact support.

Let $(\Gamma,X)$ be a measurable space and let $\mu$ and $\gamma$ be two measures on $(\Gamma,X)$. Let $\Gamma$ be a compact subset of $\mathbb{R}^n$ and denote by $t$ the ordinary Lebesgue measure on $\Gamma$. $C_0(\Gamma)$ is the space of real-valued continuous functions $f$. If $C_0(\Gamma)$ is a Banach space with the norm $$\|f\|= \sup_{x\in \Gamma}|f|,~~f\in C_0(\Gamma).$$ Show that $\Psi$ defined by $$\Psi(f):=\int_\Gamma fd\mu$$ is a bounded linear functional and $\|\Psi\| =\mu(\Gamma)$.

I have tried to read and found out that $\gamma$ is said to be absolutely continuous with respect to $\mu$ if $$\mu(E)= 0 \Rightarrow \gamma(E) = 0~~\forall E\in M.$$ However, I have failed to work out this successfully.

• Why do you ask about absolute continuity? It has little to do with the excercise. – Ranc Jul 24 '17 at 15:19
• How (or where) does $\gamma$ appear in the problem statement? And how does Lebesgue measure appear in the problem statement? – kimchi lover Jul 24 '17 at 15:20

It seems clear to me $\Psi$ is a functional, by linearity of the integral.
Also $$|\int_\Gamma fd\mu | \le \int_\Gamma |f|d\mu \le \int_\Gamma \|f\|d\mu = \|f\|\mu(\Gamma)$$ which implies that $\|\Psi\| = \mu(\Gamma)$ (take a constant 1 function for $f$ e.g.) and also $\Psi$ is bounded at the same time.
• You mean $||\Psi||=\mu (\Gamma)$ and $|\int_{\Gamma}fd\mu|$ on the left side of the inequality – Peter Melech Jul 24 '17 at 15:32