Walter Rudin Real and Complex Analysis Chapter 2

Question

Walter Rudin Real and Complex Analysis Chapter 2 2.14 Riesz representation theorem the last step.

Why did he put the absolute value of $a$ ?

Is not it sufficient to assume $f$ is positive?

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 \\
&\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) \\
&\leq \int_X f \,\mathrm{d}\mu
+ \epsilon[ 2\mu(K) + |a| + b + \epsilon ].
\end{align*}

(Original scanned image here.)

Answer 1

If you use $a$ instead of $|a|$ you cannot go from the second to the third line, since not knowing that sign of $a$ precludes you from knowing if you keep the inequality $\sum_i\Lambda_i\geq\mu(K)$.

And, if you do the proof just for $f\geq0$, you only get the inequality $\Lambda f\leq\int_Xf\,d\mu$, and not equality.