Wikinotes

Maintainer: admin

Theorem: Let $I = [a,b]$ and $J = [c,d]$ and suppose $f : I \to \mathbb{R}$ is integrable, $\varphi : J \to \mathbb{R}$ is continuous and $f(I) \subseteq J$. Then $\varphi \circ f$ is integrable on $I$.

Proof: Let $\epsilon > 0$ be given, let $k = \sup \{ |\varphi(x)| : x \in J \}$ and let $\epsilon' = \epsilon / 2(b-a+2k)$. Since $\varphi$ is uniformly continuous on $J$, there exists some $\delta > 0$ such that if $s, t \in J$ with $|s - t| < \delta$, then $|\varphi(s) - \varphi(t)| < \epsilon'$. WLOG, let $\delta < \epsilon'$. Since $f$ is integrable on $I$, there exists a partition $P$ of $I$, $P = \{ x_0, x_1, \ldots, x_n \}$ such that $U_f(P) - L_f(P) < \delta^2$ (we will show that $U_{\varphi \circ f}(P) - L_{\varphi \circ f}(P) < \epsilon$). Following our usual notation, let

$$\begin{align*} m_k &= \inf\{ f(x) : x \in [x_{k-1}, x_k] \} \\ M_k &= \sup\{ f(x) : x \in [x_{k-1}, x_k] \} \end{align*}$$

and

$$\begin{align*} \tilde{m_k} &= \inf\{ \varphi \circ f(x) : x \in [x_{k-1}, x_k] \} \\ \tilde{M_k} &= \sup\{ \varphi \circ f(x) : x \in [x_{k-1}, x_k] \} \end{align*}$$

Then

$\begin{align*} U_{\varphi \circ f}(P) - L_{\varphi \circ f}(P) &= \sum_{k=1}^n \tilde{M_k} (x_k - x_{k-1}) - \sum_{k=1}^n \tilde{m_k} (x_k - x_{k-1}) \\ &= \sum_{k=1}^n (\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) \end{align*}$

let $A = \{ k : M_k - m_k < \delta \}$ and $B = \{ k : M_k - m_k \geq \delta \}$. Observe that if $k \in A$ and $x, y \in [x_{k-1}, x_k]$ then $|f(x) - f(y)| \leq M_k - m_k < \delta$, and thus in this case

$$ |\varphi(f(x)) - \varphi(f(y))| < \epsilon' $$

i.e. if $k \in A$ then $\tilde{M_k} - \tilde{m_k} \leq \epsilon'$

and so

$\begin{align*} \sum_{k \in A} (\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) \leq \epsilon' \sum_{k \in A} x_k - x_{k-1} \leq \epsilon'(b-a) \\ \leq \epsilon'\sum_{k=1}^n x_k - x_{k-1} = \epsilon'(b-a) \end{align*}$

If $k \in B$ we know $\tilde{M_k} - \tilde{m_k} \leq 2K$ thus

$\begin{align*} \sum_{k \in B} (\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) \leq 2K\sum_{k \in B} (x_k - x_{k-1}) \end{align*}$

but $\delta \leq M_k - m_k$ so that $1 \leq \frac{M_k - m_k}{\delta}$ so

$$ 2K \sum_{k \in B}(x_k - x_{k-1}) \leq 2K \sum_{k \in B}\frac{M_k - m_k}{\delta}(x_k - x_{k-1}) \leq \frac{2K}{\delta} \sum_{k=1}^n(M_k - m_k)(x_k - x_{k-1}) = \frac{2K}{\delta}[U_f(P) - L_f(P)] < 2k\delta < 2k\epsilon' $$

Finally,

$\begin{align*} U_{\varphi \circ f}(P) - L_{\varphi \circ f}(P) &= \sum_{k=1}^n(\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) \\ &= \sum_{k\in A}(\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) + \sum_{k\in B}(\tilde{M_k} - \tilde{m_k})(x_k - x_{k-1}) \\ &\leq \epsilon'(b-a) + 2k\epsilon' = \epsilon'(b-a+2K) \\ &= \frac{\epsilon(b-a+2K)}{2(b-a+2K)} = \frac{\epsilon}{2} < \epsilon \end{align*}$

Corollary: Let $I = [a,b]$ and $f : I \to \mathbb{R}$ be integrable. Then:

(a) $|f| : I \to \mathbb{R}$ is integrable and $\left|\int_a^b f\right| \leq \int_a^b |f|$

(b) if $n \in \mathbb{N}$, $f^n$ is integrable on $I$.

Proof: $f$ is integrable and therefore understood to be bounded by $k$.

(a) Let $\varphi(c) = |x|$. $\varphi$ is continuous on $[-k,k]$ and $f(I) \subseteq [-k,k]$ so by previous theorem, $\varphi \circ f = |f|$ is integrable on $I$. Observe that $-|f| \leq f \leq |f|$ thus

$$ -\int_a^b |f| \leq \int_a^b f \leq \int_a^b |f| $$

$$ \left| \int_a^b f \; \right| \leq \int_a^b |f| $$

also $|f| \leq k$ so $\int_a^b |f| \leq \int_a^b k = k(b-a)$

(b) $\varphi(x) = x^n$ is continuous on $[-k,k]$

Monday, February 25, 2013 Integrability of function compositions

Monday, February 25, 2013
Integrability of function compositions