13 Properties of the Integral and the Fundamental Theorems
Having defined the Riemann integral, we are in a position to prove the major theorems about it, starting with some integrability conditions. The first theorem provides an easier way to show that a function is integrable. The idea is that $\ds \sup_P\{U(P,f)\} = \inf_P\{L(P,f\}$ if and only if some specific $U(Q,f)$ can be made arbitrarily close to $L(Q,f)$.
Theorem: Let $f$ be a bounded function on $[a,b]$. Then $f$ is integrable on $[a,b]$ if and only if for every $\eps>0$, there is a partition $P$ of $[a,b]$ such that $U(P,f)-L(P,f)<\eps$.
Based on this theorem, it is fairly easy to prove that certain kinds of functions are integrable:
Theorem: Let $f$ be a function defined on $[a,b]$. Then if is itegrable if it satisfies any of the following properties: $f$ is continuous, $f$ is non-decreasing, or $f$ is non-increasing. [Note that a function that satisfies any of these properties is automatically bounded.]
Note that the set of integrable functions will include some pretty crazy functions. For example, there are continuous functions that are not differentiable at any point. And there are increasing functions that have a jump discontinuity at every rational number. However, not every function is integrable. For example, Dirichlet's function, which is 1 at every irrational number and 0 at every rational number, is not integrable on any interval.
We are also in a position to prove the basic theorems about integration, leading up to the Fundamental Theorems of Calculus, which give the relationship between integration and differentiation.
Theorem (Additivity of the Integral): Suppose that $f$ is defined on $[a,b]$ and $c\in(a,b)$. Then $f$ is integrable on $[a,b]$ if and only if it is integrable on both $[a,c]$ and $[c,b]$. And if $f$ is integrable on $[a,b]$, then $$\int_a^b f = \int_a^c f + \int_c^b f$$
To remove the restriction that $a<c<b$ in this formula, we define $\int_a^bf=0$ for $a=b$, and we define $\int_a^b f = -\int_b^a f$ for $a>b$. With these definitions, it is easy to see that the additivity formula holds for any $a,b,c$ as long as $f$ is integrable on an interval that contains all three numbers.
Definition: If $f$ is a function that is defined at $a$, then we define $\ds\int_a^a f=0$. If $f$ is integrable on the interval $[b,a]$, then we define $\ds\int_a^b f=-\int_b^a f$.
The "linearity" of the integral says that the set of all integrable functions on $[a,b]$ to $\R$ is a vector space under addition of functions and multiplication of a function by a constant, and that the integral is a linear function from that vector space to $\R$.
Theorem (Linearity of the Integral): Suppose that $f$ and $g$ are integrable functions on $[a,b]$ and that $r,s\in\R$. Then the function $rf+sg$ is integrable on $[a,b]$ and $$\int_a^b(rf+sg) = r\int_a^b f\ \ +\ \ s\int_a^b g$$
Theorem (Mean Value Theorem for Integrals): If $f$ is continuous on $[a,b]$, then there is a $c\in[a,b]$ such that $$\int_a^b = f(c)\cdot(b-a)$$
The last of the major theorems about integrable functions are the Fundamental Theorems of Calculus. The first Fundamental Theorem makes it possible to find the integral of any function that has an antiderivative. The second shows that any continuous function does, in fact, have an antiderivative.
Theorem (First Fundamental Theorem of Calculus): Suppose that $f$ is integrable on $[a,b]$ and that there is function $g$ that satisfies $g'(x)=f(x)$ for all $x\in[a,b]$. (So, $g$ is an antiderivative for $f$ on $[a,b]$.) Then $$\int_a^b f = g(b)-g(a)$$
Theorem (Second Fundamental Theorem of Calculus): Suppose that $f$ is continuous on $[a,b]$. For $x\in [a,b]$, Define $$F(x) = \int_a^x f$$ Then $F'(x) = f(x)$ for all $x\in[a,b]$.
Theorem: Suppose that $f$ is integrable on $[a,b]$. For $x\in [a,b]$, Define $$F(x) = \int_a^x f$$ Then $F$ is continuous on $[a,b]$. Futhermore, if $f$ is continuous at $c$ for some $c\in[a,b]$, then $F$ is differentiable at $c$, and $F'(c)=f(c)$.
Proof: Since $f$ is integrable on $[a,b]$, it is bounded on $[a,b]$. Say, $|f(x)| \le M$ for $x\in[a,b]$. Let $c\in[a,b]$. It follows that $|F(x)-F(c)|=\big|\int_x^c f\big|\le M|x-c|$, which easily implies that $F$ is continuous at $c$.
Now, suppose that $f$ is continuous at $c$. To show $F'(c)=f(c)$, we can show $\lim_{x\to c}\frac{F(x)-F(c)}{x-c} = f(c).$ Let $\eps>0$. We need to find $\delta>0$ such that for $x\in(c-\delta,c+\delta)$, $\big|\frac{F(x)-F(c)}{x-c}-f(c)\big|<\eps.$
Since $f$ is continuous at $c$, there is a $\delta>0$ such that for $x\in(c-\delta,c+\delta)$, $|f(x)-f(c)|<\eps/2.$ Noting that $f(c)=\frac{1}{x-c}\int_x^c f(c)\,dt$, we can calculate that for $x\in(c-\delta,c+\delta),$
\begin{align*} \left|\frac{F(x)-F(c)}{x-c}-f(c)\right| &=\left|\frac{1}{x-c}\int_x^c f(t)\,dt - \frac{1}{x-c}\int_x^c f(c)\,dt\right|\\ &=\frac{1}{|x-c|}\cdot\left| \int_x^c (f(t)-f(c))\,dt \right|\\ &\le\frac{1}{|x-c|}\cdot\int_x^c\big|f(t)-f(c)\big|\,dt\\ &\le\frac{1}{|x-c|}\cdot\int_x^c \eps/2\,dt\\ &=\frac{1}{|x-c|}\cdot |x-c|\eps/2\\ &=\eps/2\\ &<\eps \end{align*}