# Wednesday, February 20, 2013 Additivity Theorem for integrals

Recall last class we covered the following theorem:

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

Corollary: Let $I = [a,b]$, $f, g | I \to \mathbb R$ integrable on $I$ such that $f(x) \leq g(x)$ for all $x \in I$. Then

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

Proof: Let $h = g - f$. Then $h : I \to \mathbb R$ is integrable and $h \geq 0$ for all $x \in I$. Then $\int_a^b h \geq 0$, but

$$\int_a^b h = \int_a^b g - f = \int_a^b g - \int_a^b f$$

Thus $\int_a^b f \leq \int_a^b g$.

Theorem: Let $I = [a,b]$, $c \in (a,b)$ and $f : I \to \mathbb R$ a bounded function on $I$. Then $f$ is integrable on $I$ $\iff$ $f$ is integrable on $[a,c]$ and on $[c,b]$. In this case,

$$\int_a^b f = \int_a^c f + \int_c^b$$

Proof: ($\Leftarrow$) For $\epsilon > 0$ given, there exist partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$ such that

$$U_f(P_1) - L_f(P_1) < \frac{\epsilon}{2} \text{ and } U_f(P_2) - L_f(P_2) < \frac{\epsilon}{2}$$

Let $P = P_1 \cup P_2$. $U_f(P) = U_f(P_1) + U_f(P_2)$ and $L_f(P) = L_f(P_1) + L_f(P_2)$ so that

\begin{align*} U_f(P) - L_f(P) &= U_f(P_1) + U_f(P_2) - L_f(P_1) - L_f(P_2) \\ &= U_f(P_1) - L_f(P_1) + U_f(P_2) - L_f(P_2) \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*}

so $f$ is integrable on $[a,b]$ by Riemann Criterion.

($\Rightarrow$) Let $f$ be integrable on $I$. Then for $\epsilon > 0$ given, there exists a partition $P$ of $I$ such that $U_f(P) - L_f(P) < \epsilon$. Let $P_c = P \cup \{c\}$, a refinement of $P$. So $U_f(P_c) - L_f(P_c) < \epsilon$. Let $P_1 = P_c \cap [a,c]$ and $P_2 = P_c \cap [c,b]$. Then $U_f(P_c) = U_f(P_1) + U_f(P_2)$ and $L_f(P_c) = L_f(P_1) + L_f(P_2)$, i.e.

\begin{align*} U_f(P_c) - L_f(P_c) &= U_f(P_1) + U_f(P_2) - L_f(P_1) - L_f(P_2) \\ &= U_f(P_1) - L_f(P_1) + U_f(P_2) - L_f(P_2) < \epsilon \end{align*}

Since $U_f(P_1) - L_f(P_1) \geq 0$ and $U_f(P_2) - L_f(P_2) \geq 0$ we conclude that $U_f(P_1) - L_f(P_1) < \epsilon$ and $U_f(P_2) - L_f(P_2) < \epsilon$.

Thus $f$ is integrable on $[a,c]$ and $[c,b]$. Now we must show that $\int_a^b f = \int_a^c f + \int_c^b f$.

\begin{align*} \int_a^b f = U(f) \leq U_f(P_c) &= U_f(P_1) + U_f(P_2) \\ &< L_f(P_1) + L_f(P_2) + \epsilon \\ &\leq L_{[a,c]}(f) + L_{[c,b]}(f) \\ &= \int_a^c f + \int_c^b f + \epsilon \end{align*}

Since $\epsilon > 0$ arbitrary, $\int_a^b f \leq \int_a^c f + \int_c^b f$. We can use a similar argument to show the $\int_a^b f \geq \int_a^c f + \int_c^b f$ from which we can conclude that $\int_a^b f = \int_a^c f + \int_c^b f$.

Remark: Of course, you can easily extend this by induction.