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The assumption of continuity of \(f\) at \(x_{0}\) gives us control over the difference \(|f(t)-f(x_{0})|\), i.e., given \(\epsilon>0\), there is a \(\delta>0\) such that
\[|x-x_{0}|<\delta\quad\text{implies}\quad|f(x)-f(x_{0})|<\epsilon.\]
Therefore,
\[\left|\frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0})\right|<\epsilon\quad\text{if} \quad 0<|x-x_{0}|<\delta,\]
which proves that \(F\) is differentiable at \(x_{0}\) and that \(F^{\prime}(x_{0})=f(x_{0})\).
Note that the conclusion of the above theorem can be expressed using Leibnitz’s notation as
\[\frac{d}{dx}\left(\int_{a}^{x}f(t)\,dt\right)=f(x).\]
For example, let \(F(x)=\int_{0}^{x}\sqrt{t^{2}+3}\;dt\), for \(x\geq 0\), then \(F^{\prime}(x)=\sqrt{x^{2}+3}\) or
\[\frac{d}{dx}\left(\int_{0}^{x}\sqrt{t^{2}+3}\,dt\right)=\sqrt{x^{2}+3}.\]
**Example 51**.: Let \(F(x)=\int_{0}^{x^{2}}\sqrt{t^{4}+3}\ dt\), for \(x\geq 0\), and suppose we want to find \(F^{\prime}(x)\) using the Fundamental Theorem of Calculus I. In this problem \(F(x)=\int_{a}^{g(x)}f\,dt\), where \(g(x)=x^{2}\). Thus \(F=G\circ g\), where \(G(x)=\int_{a}^{x}f\,dt\), and we can use the Chain Rule to write
\[F^{\prime}(x)=G^{\prime}(g(x))g^{\prime}(x),\]
where \(g^{\prime}(x)=2x\) and \(G^{\prime}(g(x))=\sqrt{(g(x))^{4}+3}\). Thus
\[F^{\prime}(x)=\left(\sqrt{(x^{2})^{4}+3}\right)(2x).\]
**Theorem 52** (Fundamental Theorem of Calculus II).: _If \(f\) is differentiable on \([a,b]\) and \(f^{\prime}\) is integrable on \([a,b]\), then_
\[\int_{a}^{b}f^{\prime}=f(b)-f(a).\]
Proof
: Let \(P\) be a partition of \([a,b]\). By applying the Mean Value Theorem to each subinterval \([x_{k-1},x_{k}]\), we obtain points \(t_{k}\in(x_{k-1},x_{k})\) such that
\[f(x_{k})-f(x_{k-1})=f^{\prime}(t_{k})(x_{k-1}-x_{k}).\]
Thus we have
\[f(b)-f(a)=\sum_{k=1}^{n}[f(x_{k})-f(x_{k-1})]=\sum_{k=1}^{n}f^{\prime}(t_{k})( x_{k-1}-x_{k}),\]
and it follows that
\[L(f^{\prime},P)\leq f(b)-f(a)\leq U(f^{\prime},P).\]
Since this holds for each partition \(P\), we also have
\[L(f^{\prime})\leq f(b)-f(a)\leq U(f^{\prime}).\]
But \(f^{\prime}\) is assumed to be integrable on \([a,b]\), thus, \(L(f^{\prime})=U(f^{\prime})=\int_{a}^{b}f^{\prime}\,dx\) and \(\int_{a}^{b}f^{\prime}\,dx=f(b)-f(a)\).
The Fundamental Theorem of Calculus II provides the standard device of anti-differentiation to compute the integral. For example, if we let \(f(x)=x^{3}\) for \(x\in\mathbb{R}\), then \(f^{\prime}(x)=3x^{2}\), and
\[\int_{1}^{2}3x^{2}\ dx=f(2)-f(1)=8-1=7.\]
It is also the basis for a well-known formula for _integration by parts_ which we will give in the following example.
**Example 52** (Integration by Parts).: Suppose that \(f\) and \(g\) are differentiable on \([a,b]\) and that \(f^{\prime}\) and \(g^{\prime}\) are integrable on \([a,b]\). Then the following integration by parts formula holds:
\[\int_{a}^{b}(fg^{\prime})\,dx=[f(b)g(b)-f(a)g(a)]-\int_{a}^{b}(f^{\prime}g)\,dx.\]
To see this let \(h=fg\), then \(h^{\prime}=f^{\prime}g+fg^{\prime}\). Now both \(f\) and \(g\) are differentiable, hence continuous and therefore integrable on \([a,b]\). Thus \(h^{\prime}\) is integrable on \([a,b]\), and from the Fundamental Theorem of Calculus II, we obtain
\[\int_{a}^{b}h^{\prime}\,dx=h(b)-h(a),\]
that is,
\[\int_{a}^{b}[f^{\prime}g+fg^{\prime}]\,dx=\int_{a}^{b}(f^{\prime}g)\,dx+\int_{ a}^{b}(fg^{\prime})\,dx=f(b)g(b)-f(a)g(a).\]
### Sets of Measure Zero and Lebesgue’s Criterion
When we constructed the Cantor set, we saw that it has zero length, and we stated this as “the Cantor set has measure zero.” Now we give a precise definition to this concept.
**Definition 38**.: A set \(A\subseteq\mathbb{R}\) is said to be _measure zero_ if, for every \(\epsilon>0\), there exists a countable collection of open intervals \(O_{n}\) such that
\[A\subseteq\bigcup_{n=1}^{\infty}O_{n}\quad\text{and}\quad\sum_{n=1}^{\infty}|O _{n}|\leq\epsilon,\]
where by \(|O_{n}|\) we mean the length of the interval \(O_{n}\).
It is clear that if \(A\) is a set of measure zero and \(K\subset A\), then \(K\) has measure zero as well.
**Example 53**.:
1. A single point \(A=\{a\}\) is a set of measure zero. For \(n=1\), let \(O_{1}=(a-\epsilon,a+\epsilon)\), and for \(n\geq 2\), set \(O_{n}=\emptyset\); then \(\bigcup_{n=1}^{\infty}O_{n}\) covers \(A\) and the intervals have the total length \(\epsilon\).
2. A finite set \(A=\{a_{1},a_{2},\ldots,a_{k}\}\) has measure zero. Let \(\epsilon>0\) be arbitrary, and for each \(1\leq n\leq k\), consider open sets of the form \(O_{n}=\Big{(}a_{n}-\frac{\epsilon}{2k},a_{n}+\frac{\epsilon}{2k}\Big{)}\). Then \[A\subseteq\bigcup_{n=1}^{k}O_{n}\quad\text{and}\quad\sum_{n=1}^{k}|O_{n}|=\sum _{n=1}^{k}\frac{\epsilon}{k}=\epsilon.\]* If \(A\) is a countable subset of \(\mathbb{R}\), then \(A\) is a set of measure zero. Let \(A=\{a_{1},a_{2},\dots\}\) be a countable infinite set. Let \(\epsilon>0\) be arbitrary and for \(i\in\mathbb{N}\); set \[O_{i}=\left(a_{i}-\frac{\epsilon}{2^{i+1}},a_{i}+\frac{\epsilon}{2^{i+1}} \right).\] Observe that \(a_{i}\in O_{i}\) and \(|O_{i}|=\epsilon 2^{-i}\) for \(i\in\mathbb{N}\). Therefore, \[A\subseteq\bigcup_{i=1}^{\infty}O_{i}\quad\text{and}\quad\sum_{n=1}^{\infty}| O_{i}|=\epsilon\sum_{n=1}^{\infty}\frac{1}{2^{i}}=\epsilon.\]
* \(\mathbb{Q}\) has measure zero since it is countable.
* If \(A_{1},A_{2},\dots\) is a sequence of sets of measure zero, then \(A=\bigcup_{n=1}^{\infty}A_{n}\) is also a set of measure zero (try to prove it).
* The Cantor set \(C\) has measure zero. Recall that the Cantor set \(C\) is uncountable, but for each \(n\), the Cantor set is contained in a finite union of intervals of total length \(\frac{2^{n}}{3^{n}}\) and (measure of \(C\)) \(\leq\frac{2^{n}}{3^{n}}\to 0\).
Above we proved that continuous functions are Riemann integrable and saw examples of functions with only a finite number of discontinuities that are also integrable. However the following example illustrates the fact that the set of discontinuities of an integrable function can be infinite, even uncountable.
**Example 54**.: Let \(0 \[f(x)=\left\{\begin{array}{ll}\frac{1}{q}&\text{if $x\in[a,b]\cap\mathbb{Q}$ and}x=\frac{p}{q}\text{ where $p$ and $q$ are coprime;}\\ \\ 0&\text{if $x\in[a,b]$ is irrational.}\end{array}\right.\] This function is continuous at each irrational number in \((0,1)\) and discontinuous at each rational number in \((0,1)\). We claim that \(f(x)\) is Riemann integrable over \([a,b]\) and that \(\int_{a}^{b}f(x)dx=0\). Let \(\mathcal{P}\) be a partition of \([a,b]\). If \(\mathcal{P}=\{x_{0}=a \[\inf\{f(x):x_{i} because \(\mathbb{R}\setminus\mathbb{Q}\) is dense in \(\mathbb{R}\). Hence we have \(L(\mathcal{P},f)=0\), which implies \[L(f)=\sup L(\mathcal{P},f)=0,\] where the supremum is taken over all partitions of \([a,b]\). Let us now show that \(U(f)=L(f)=0\). Fix \(\epsilon>0\). Then the set \[B_{\epsilon}=\left\{\frac{p}{q}\in[a,b]\cap\mathbb{Q};\text{ where $p$ and $q$ are coprime and }\frac{1}{q}\geq\frac{\epsilon}{2(b-a)}\right\}\]is finite. Without loss of generality, assume that \(B_{\epsilon}\) is not empty and has \(n\geq 1\) elements. Set
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