This is the definition which we need for the theorem:
There is the theorem:
Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a
positive linear functional on $C_{c}(X)$. Then there exists a $\sigma$-algebra $\mathfrak{M}$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathfrak{M}$ which represents $\Lambda$ in the sense that:
$(a)$ $\Lambda f$ $=$ $\int_X f d\mu $ for every $f$ $\in$ $C_{c}(X)$.
(The proof of this theorem is divided in $10$ steps. )
Step $X$ :
For every $f$ $\in$ $C_{c}(X)$, $\Lambda f $ $=$ $\int_X f d\mu $
This proves (a), and completes the theorem.
Proof.
Clearly, it is enough to prove this for real $f$.
Also, it is enough to prove the inequality
\begin{equation}
\tag{16}
\Lambda f
\leq \int_X f \,\mathrm{d}\mu
\end{equation}
for every real $f \in C_c(X)$.
For once $(16)$ is established, the linearity of $\Lambda$ shows that
$$
-\Lambda f
= \Lambda(-f)
\leq \int_X (-f) \,\mathrm{d}\mu
= – \int_X f \,\mathrm{d}\mu,
$$
which, together with $(16)$ shows that equality holds in $(16)$.
Let $K$ be the support of the real $f \in C_c(X)$, let $[a,b]$ be an interval which contains the range of $f$ (note the Corollary to Theorem 2.10), choose $\epsilon > 0$, and choose $y_i$, for $i = 0, 1, \dotsc, n$, so that $y_i – y_{i-1} < \epsilon$ and
\begin{equation}
\tag{17}
y_0 < a < y_1 < \dotsb < y_n = b.
\end{equation}
Put
\begin{equation}
\tag{18}
E_i
= \{ x : y_{i-1} < f(x) \leq y_i \} \cap K
\qquad
(i = 1, \dotsc, n)
\end{equation}
Since $f$ is continuous, $f$ is Borel measurable, and the sets $E_i$ are therefore disjoint Borel sets whose union is $K$.
There are open sets $V_i \supset E_i$ such that
\begin{equation}
\tag{19}
\mu(V_i)
< \mu(E_i) + \frac{\epsilon}{n}
\qquad
(i = 1, \dotsc, n)
\end{equation}
and such that $f(x) < y_i + \epsilon$ for all $x \in V_i$.
By Theorem 2.13, there are functions $h_i \prec V_i$ such that $\sum h_i = 1$ on $K$.
Hence $f = \sum h_i f$, and Step II shows that
$$
\mu(K)
\leq \Lambda\left( \sum h_i \right)
= \sum \Lambda h_i.
$$
Since $h_i f \leq (y_i + \epsilon) h_i$, and since $y_i – \epsilon < f(x)$ on $E_i$, we have
\begin{align*}
\Lambda f
&= \sum_{i=1}^n \Lambda(h_i f)
\leq \sum_{i=1}^n (y_i + \epsilon) \Lambda h_i \\
&= \sum_{i=1}^n (|a| + y_i + \epsilon) \Lambda h_i
– |a| \sum_{i=1}^n \Lambda h_i \\
(1) &\leq \sum_{i=1}^n (|a| + y_i + \epsilon)[ \mu(E_i) + \epsilon/n ]
– |a| \mu(K) \\
&= \sum_{i=1}^n (y _i – \epsilon) \mu(E_i)
+ 2 \epsilon \mu(K)
+ \frac{\epsilon}{n} \sum_{i=1}^n (|a| + y_i + \epsilon) \\
(2) &\leq \int_X f \,\mathrm{d}\mu
+ \epsilon[ 2\mu(K) + |a| + b + \epsilon ].
\end{align*}
I don't understand why it is enough to prove this for real $f$.
I also don't understand why is it enough to prove the inequality.
$\Lambda f $ $\leq$ $\int_X f d\mu$ ( we have to show equality).
Any help would be appreciated.
Best Answer
It is enough to prove this for real $f$, since $\Lambda$ and integration are both $\mathbb{C}$-linear. That is, write $f = u+iv$ for $u, v$ real and see that if this is proved for real functions, $$\Lambda (u + iv) = \Lambda u + i \Lambda v = \int_X u d\mu + i \int_X v d\mu = \int_X u + iv d\mu.$$
It is enough to prove the inequality because if $A, B$ are real numbers and $A \leq B$ and $-A \leq -B$ then $A = B$ (just reorganize the second inequality to see $A \geq B$). Apply this with $A = \Lambda f, B = \int_X f d\mu$.