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The following theorem is from Rudin's $\textit{Principles of Mathematical Analysis}$.

$\textbf{Theorem:}$ Let $\alpha$ be monotonically increasing on $[a,b]$. Suppose $f_n \in \mathscr{R} (\alpha)$ on $[a,b]$, for $n = 1,2,3,...,$ and suppose $f_n \rightarrow f$ uniformly on $[a,b]$. Then $f \in \mathscr{R}(\alpha)$ on $[a,b]$, and $$\int_{a}^{b} f d\alpha = \lim_{n \rightarrow \infty} \int_{a}^{b} f_n d \alpha. \tag{*}$$

$\textbf{Proof}:$ Put $$\epsilon_n = \sup_{a \leq x \leq b}|f_n (x) - f(x)|.$$ Then $$f_n - \epsilon_n \leq f \leq f_n + \epsilon_n,$$ so that the upper and lower integrals of $f$ satisfy \begin{equation}\int_a^b (f_n - \epsilon_n) d\alpha \leq \int_{\_} f d \alpha \leq \int^{\_} f d \alpha \leq \int_a^b (f_n + \epsilon_n) d \alpha. \tag{1}\end{equation}

Hence

$$0 \leq \int^{\_} f d\alpha - \int_{\_} f d \alpha \leq 2 \epsilon_n [\alpha (b) - \alpha (a) ] \tag{2}.$$ Since $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$, the upper and lower integrals of $f$ are equal.

Thus $f \in \mathscr{R}(\alpha)$. Another application of $(1)$ now yields $$\left| \int_a^b f d \alpha - \int_a^b f_n d \alpha \right| \leq \epsilon_n [\alpha(b) - \alpha(a)], \tag{3}$$ which implies the result $(*)$.

$\textbf{My question:}$ Where do lines $(2)$ and $(3)$ come from? I think line $(2)$ just comes from rearranging line $(1)$ via basic operations, but I don't see where the integrals went.

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(1) consists of 3 inequalities:

$$\begin{equation}\int_a^b (f_n - \epsilon_n) d\alpha \leq \int_{\_} f d \alpha \tag{1.1}\end{equation}$$

$$\begin{equation}\int_{\_} f d \alpha \leq \int^{\_} f d \alpha \tag{1.2}\end{equation}$$

$$\begin{equation}\int^{\_} f d \alpha \leq \int_a^b (f_n + \epsilon_n) d \alpha. \tag{1.3}\end{equation}$$

(1.2) gives you

$$0 \leq \int^{\_} f d\alpha - \int_{\_} f d \alpha$$

with (1.1) and (1.3) you get:

$$\begin{align*}\int^{\_} f d\alpha - \int_{\_} f d \alpha &\le \int_a^b (f_n + \epsilon_n) d \alpha - \int_a^b (f_n - \epsilon_n) d\alpha \\ &= 2\varepsilon_n\int_a^b 1\, d\alpha \\ &= 2\varepsilon_n[\alpha(b) - \alpha(a)]\end{align*}$$

(3) you get by similar steps…

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  • $\begingroup$ This is simple to follow. Thanks. $\endgroup$ – amarney Mar 21 '17 at 13:53

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