Walter Rudin Real and Complex Analysis Chapter 2 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.)


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. 
