# 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 $$\tag{16} \Lambda f \leq \int_X f \,\mathrm{d}\mu$$ 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 $$\tag{17} y_0 < a < y_1 < \dotsb < y_n = b.$$ Put $$\tag{18} E_i = \{ x : y_{i-1} < f(x) \leq y_i \} \cap K \qquad (i = 1, \dotsc, n)$$ 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 $$\tag{19} \mu(V_i) < \mu(E_i) + \frac{\epsilon}{n} \qquad (i = 1, \dotsc, n)$$ 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.)

• Since every real function $f$ can be written $f=f^+ - f^-$ both of them are positive Aug 24, 2018 at 22:50
• To avoid putting the absolute value $a$ Aug 24, 2018 at 22:54
• Why can't I apply what I said to this proof ? Aug 25, 2018 at 2:28

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.