Monday, February 18, 2013

Linearity of the integral

Theorem: Let $I = [a,b]$ and $f,g : I \to \mathbb{R}$ be Riemann integrable on $I$. Then for $k \in \mathbb{R}$, $kf$ and $f + g$ are Riemann integrable on $I$ and

$$ \int_a^b kf = k \int_a^b f \quad \text{and} \quad \int_a^b f + g = \int_z^b f + \int_a^b g $$

Proof: Trivial if $k = 0$. Assume $k < 0$. Let $P = \{x_0, x_1, \ldots, x_n\}$ be a partition of $I$. Since $k < 0$,

$$ \inf\{kf(x) : x \in [x_{j-1}, x_j]\} = k\cdot\sup\{f(x) : x \in [x_{j-1}, x_j]\} $$

Thus if we multiply by $(x_j - x_{j-1})$ and add these up, we can conclude that $L_{kf}(P) = kU_f(P)$. Again, since $k < 0$,

$$\begin{align} &\sup\{L_{kf}(P) : P\text{ is a partition of }I\} \\ =&\sup\{kU_f(P) : P\text{ is a partition of }I\} \\ =&k\cdot\inf\{U_f(P) : P\text{ is a partition of }I\} \end{align}$$

i.e. $L(kf) = kU(f)$.

Similarly, we can show that $U(kf) = kL(f)$. As $f$ is Riemann integrable, $L(f) = U(f)$. Thus $L(kf) = U(kf)$ and $kf$ is Riemann integrable. In fact, $L(kf) = \int_a^b kf = kU(f) = k\int_a^b f$.

For the claim concerning $f + g$, recall that

$$ \inf\{f(x) : x \in [x_{j-1}, x_j]\} + \inf\{g(x) : x \in [x_{j-1}, x_j]\} \leq \inf\{f(x) + g(x) : x \in [x_{j-1}, x_j]\} $$

And similarly

$$ \sup\{f(x) : x \in [x_{j-1}, x_j]\} + \sup\{g(x) : x \in [x_{j-1}, x_j]\} \geq \sup\{f(x) + g(x) : x \in [x_{j-1}, x_j]\} $$

Thus it follows that $L_f(P) + L_g(P) \leq L_{f+g}(P)$ and $U_f(P) + U_g(P) \geq U_{f+g}(P)$. Let $\epsilon > 0$. Since $f$ and $g$ are integrable, by the Riemann criterion there exists some $P_{f, \epsilon}$ and $P_{g,\epsilon}$, partitions of $I$, such that

$$ U_f(P_{f,\epsilon}) < L_f(P_{f,\epsilon}) + \frac{\epsilon}{2} $$

and

$$ U_g(P_{g,\epsilon}) < U_f(P_{g,\epsilon}) + \frac{\epsilon}{2} $$

Let $P_\epsilon = P_{f,\epsilon} \cup P_{g, \epsilon}$, a refinement of both $P_{f, \epsilon}$ and $P_{g, \epsilon}$.

Then $U_f(P_\epsilon) < L_f(P_\epsilon) + \frac{\epsilon}{2}$ and $U_g(P_\epsilon) < L_g(P_\epsilon) + \frac{\epsilon}{2}$ so that

$$ U_{f+g}(P_\epsilon) \leq U_f(P_\epsilon) + U_g(P_\epsilon) < L_f(P_\epsilon) + \frac{\epsilon}{2} + L_g(P_\epsilon) + \frac{\epsilon}{2} \leq L_{f+g}(P_\epsilon) + \epsilon $$

i.e. $U_{f+g}(P_\epsilon) < L_{f+g}(P_\epsilon) + \epsilon \implies U_{f+g}(P_\epsilon) - L_{f+g}(P_\epsilon) < \epsilon$.

Thus $f+g$ is Riemann integrable by the Riemann Criterion. Observe that

$$ \begin{align*} \int_a^b f + g &= U(f+g) \\ &\leq U_{f+g}(P_\epsilon) \\ &\leq U_f(P_\epsilon) + U_g(P_\epsilon) \\ &\leq L_f(P_\epsilon) + L_g(P_\epsilon) + \epsilon \\ &\leq L(f) + L(g) + \epsilon \\ &= \int_a^b f + \int_a^b g \quad + \epsilon \end{align*} $$

For $\epsilon > 0$ arbitrary, we conclude that

$$ \int_a^b f + g \leq \int_a^b f + \int_a^b g $$

Similarly

$$ \begin{align*} \int_a^b f + \int_a^b g &= U(f) + U(g) \\ &\leq U_f(P_\epsilon) + U_g(P_\epsilon) \\ &< L_f(P_\epsilon) + L_g(P_\epsilon) + \epsilon \\ &\leq L_{f+g}(P_\epsilon) + \epsilon \\ &\leq L(f+g) + \epsilon \\ &= \int_a^b f + g \quad+ \epsilon \end{align*} $$

So $\int_a^b f + \int_a^b g \leq \int_a^b f + g$ and so

$$ \int_a^b f + g = \int_a^b f + \int_a^b g $$

Theorem: Let $I = [a,b]$ and $f : I \to \mathbb{R}$ be integrable on $I$. If $f \geq 0$ on $I$ then $\int_a^b f \geq 0$.

Proof: Recall that if $m \leq f(c) \leq M$ on $I$, then $m(b-a) \leq \int_a^b f \leq M(b-a)$. In this case, $m \geq 0$. Thus, we're done.

Corollary: If $I = [a,b]$ and $f : I \to \mathbb{R}$ is Riemann integrable, then if $f$ is bounded by $k$ on $I$,

$$ \left| \int_a^b f \right| \leq k(b-a) $$

Proof: $|f(x)| \leq k \iff -k \leq f(c) \leq k$. By previous observations, $-k(b-a) \leq \int_a^b f \leq k(b-a)$ so $|\int_a^b f| \leq k(b-a)$.