Proof. To prove (i) it suffices to assume that \(P^*\) contains exactly one more point than \(P\), say the point \(c\). Let \(c\) be in the \(j^{\text {th}}\) subinterval of \(P\). Then \[U\big (P^*, f\big ) = \sum ^n_{\substack {k = 1\\ k\neq j}}M_k(f)\, \Delta x_k + M'\, (c - x_{k - 1}) + M''(x_k - c)\] where \(M'\) and \(M''\) denote the supremum of \(f\) in \([x_{k - 1}, c]\) and \([c,x_k]\) respectively. But since \(M'\leq M_i(f)\) and \(M'' \leq M_i(f)\) we have \(\, U\big (P^*, f\big ) \leq U\big (P,f\big )\).
The inequality for the lower sum is proved in a similar way.

To prove (ii) Let \(\, P= P_1\cup P_2\). Then we have \[L\big (P_1,f\big ) \leq L\big (P,f\big ) \leq U\big (P,f\big ) \leq U\big (P_2,f\big )\] \([\)the lower sum of the refinement can not exceed that of the partition\(]\).

\begin {align*} M & = \sup \big \{f(x):\, x\in [a,b]\big \}\\ m & = \inf \big \{f(x):\, x\in [a,b]\big \} \end {align*}

and \(P_1\) and \(P_2\) are two partitions of \([a,b]\), then \(\, m(b-a)\leq L\big (P_1,f\big ) \leq U\big (P_2,f\big )\leq M(b-a)\)

□

Proof. Let \(\varepsilon > 0\) be given. Then there is a partition \(P_1\) such that \[U\big (P_1,f \big ) < \overline {\int ^b_a} f + \varepsilon \] It follows that \(\, \displaystyle {\overline {\int ^b_a} f + \varepsilon }\,\) is an upper bound to all lower sum \(L\big (P_1,f\big )\). Hence \[\underline {\int ^b_a} f \leq \overline {\int ^b_a} f + \varepsilon \] since \(\varepsilon \) is arbitrary, this implies that \[\, \displaystyle {\underline {\int ^b_a} f \leq \overline {\int ^b_a} f}\] It is possible to have a strict inequality \[\, \displaystyle {\underline {\int ^b_a} f < \overline {\int ^b_a} f}\]

□

Proof. Let \(\, P:\, a = < x_0 < x_i < \cdots \cdots \cdots < x_k\,\cdots \cdots \, x_{n - 1} < x_n = b\, \) be any partition, then \begin {align*} U\big (P,f\big ) & = \sum ^n_{k = 1} M_k(f)\, \Delta x_k = \sum ^n_{k = 1} C\, \Delta x_k\\\\ & = C\, (b - a) \end {align*}

\begin {align*} L\big (P,f\big ) & = \sum ^n_{k = 1} m_k (f) \, \Delta x_k = \sum ^n_{k = 1} C\, \Delta x_k\\\\ & = C\, (b - a) \end {align*}

Since the partition \(P\) was arbitrary, we have \(U\big (P,f\big ) = L\big (P,f\big ) = C\, (b- a)\,\) for every partition \(P\) of \([a,b]\). Thus \[\overline {\int ^b_a} f = C\, (b - a) = \underline {\int ^b_a} f\]

Therefore, \(f\) is Riemann integrable and \[\, \displaystyle {\int ^b_a f(x)\, dx = C\, (b - a)}\]

□

Solution. The component intervals of the partition \(P_n\) are \(\, I_1 = \big [0\,,\,1/n\big ]\, , \hspace {0.3cm} I_2 = \big [1/n\,,\, 2/n\big ]\, , \hspace {0.3cm}\cdots \cdots \cdots \,,\\ \hspace {0.3cm} I_n = \big [(n - 1)/n\, ,\, n/n\big ]\). So that \(\, \Delta x_k = 1/n\,\) for \(\, k = 1,\, 2, \, \cdots \cdots \, , n\) \[M_1(f) = \dfrac {1}{n^2}\hspace {0.3cm} , \hspace {0.3cm} M_2(f)= \Bigg (\dfrac {2}{n}\Bigg )^2\hspace {0.3cm},\, \cdots \cdots \cdots \, \hspace {0.3cm} M_k(f) = \Bigg (\dfrac {k}{n}\Bigg )^2\hspace {0.3cm},\hspace {0.3cm} M_n(f) = \Bigg (\dfrac {n}{n}\Bigg )^2\] Also \[m_1(f) = 0\hspace {0.3cm},\hspace {0.3cm}m_2(f)= \Bigg (\dfrac {1}{n}\Bigg )^2\hspace {0.3cm},\hspace {0.3cm} m_3(f) = \Bigg (\dfrac {2}{n}\Bigg )^2\hspace {0.3cm},\, \cdots \cdots \, , \hspace {0.3cm} m_k (f) = \Bigg (\dfrac {k - 1}{n}\Bigg )^2\hspace {0.3cm},\hspace {0.3cm}\] \[ m_n(f) = \Bigg (\dfrac {n - 1}{n}\Bigg )^2\]

\begin {align*} U\big (P_n,f\big ) & = \sum ^n_{k = 1} M_k(f)\, \Delta x_k = \sum ^n_{k = 1}\Bigg (\dfrac {k}{n}\Bigg )^2\, \cdot \, \dfrac {1}{2}\\\\ & = \frac {1}{n^2}\, \cdot \, \dfrac {1}{n}\, + \, \Bigg (\dfrac {2}{n}\Bigg )^2\cdot \frac {1}{n}\, + \, \Bigg (\frac {3}{n}\Bigg )^2\cdot \frac {1}{n} \, + \, \cdots \cdots \, + \, \Bigg (\dfrac {n}{n}\Bigg )^2\, \cdot \, \dfrac {1}{n}\\\\ & = \dfrac {1}{n^3}\, \Big (1 + 2^2 + 3^2 + \cdots \cdots \cdots + n^2\Big )\\\\ & = \frac {1}{n^3}\, \Bigg (\dfrac {n\,(n + 1)\, (2n + 1)}{6}\Bigg )\\\\ & = \dfrac {n\, (n+1)\, (2n + 1)}{6\, n^3}\\\\ \therefore \hspace {0.3cm} U\big (P_n,f\big ) & = \dfrac {\dfrac {n}{n}\, \Big (\dfrac {n + 1}{n}\Big )\, \Big (\dfrac {2n + 1}{n}\Big )}{6} \end {align*}

Taking the limit as \(\, n \longrightarrow \infty \) \[\lim _{n\rightarrow \infty }\,U\big (P_n,f\big ) = \lim _{n\rightarrow \infty }\,\dfrac {\dfrac {n}{n}\, \Big (\dfrac {n + 1}{n}\Big )\, \Big (\dfrac {2n + 1}{n}\Big )}{6} = \frac {1\, (2)}{6} = \frac {1}{3}\]

Similarly, \begin {align*} L\big (P_n,f\big ) & = \sum ^n_{k = 1}m_k(f)\, \Delta x_k = \sum ^n_{k = 1}\Bigg (\frac {k - 1}{n}\Bigg )^2\,\cdot \,\dfrac {1}{n}\\\\ & = \frac {0}{n^2}\,\cdot \, \dfrac {1}{n}\, + \, \frac {1}{n^2}\, \cdot \, \frac {1}{n}\, + \, \Bigg (\frac {2}{n}\Bigg )^2\,\cdot \, \frac {1}{n} \, + \, \cdots \cdots \cdots \, + \, \Bigg (\frac {n - 1}{n}\Bigg )^2\,\cdot \,\frac {1}{n}\\\\ & = \frac {1}{n^3}\,\,\Big ( 0 + 1^2 + 2^2 + 3^2 + \,\cdots \cdots \cdots \, + \big (n - 1\big )^2\Big )\\\\ & = \frac {1}{n^3}\,\,\Big (1^2 + 2^2 + 3^2 + \,\cdots \cdots \cdots \, + \, \big (n - 1\big )^2\Big )\\\\ & = \frac {1}{n^3}\, \,\Bigg (\dfrac {n\, (n + 1)\, (2n + 1)}{6}\Bigg )\\\\ & = \frac {n\, (n + 1) \, (2n + 1)}{6\, n^3}\\\\ \end {align*}

Proof. We already have seen that \(\, \displaystyle {\underline {\int ^b_a} f(x)\, dx \, \leq \, \overline {\int ^b_a} f(x)\, dx}\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (1)\)
Suppose now that \(\varepsilon > 0\) is such that there is a partition \(P\) for which \(\, U\big (P,f\big ) - L\big (P, f\big ) < \varepsilon \,\) since \(\, \displaystyle {\overline {\int ^b_a} f(x) \, dx \, \leq \, U\big (P,f\big )}\,\) and \(\, \displaystyle {\underline {\int ^b_a} f(x)\,dx\, \geq \, L\big (P,f\big )}\), then we have \(\,\displaystyle {\overline {\int ^b_a}f(x)\, dx\, -\, \underline {\int ^b_a}f(x)\, dx}\, < \varepsilon \) Since \(\varepsilon > 0\) is arbitrary, we get \[\overline {\int ^b_a}f(x)\, dx \, \leq \, \underline {\int ^b_a}f(x)\, dx \hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (2)\] Hence from (1) and (2) we get \[\underline {\int ^b_a}f(x)\, dx\, = \, \overline {\int ^b_a}f(x)\, dx\] So that \(f\) is Riemann integrable on \([a,b]\).

Conversely, suppose that \(f\) is Riemann integrable on \([a,b]\). Then we have \begin {align*} \overline {\int ^b_a}f(x)\, dx \, & = \inf _P\, U\big (P_1,f\big )\\\\ & = \sup _P\, L\big (P_2,f\big )\\\\ & = \underline {\int ^b_a}f(x)\, dx \end {align*}

Let \(\varepsilon > 0\) be given, then we can choose partitions \(P_1\) and \(P_2\) such that \(\, \displaystyle {U\big (P_1,f\big ) < \overline {\int ^b_a}f(x)\, dx \, + \, \frac {\varepsilon }{2}}\,\) and \(\, \displaystyle {L\big (P_2,f\big ) > \underline {\int ^b_a}f(x)\, dx\, - \, \frac {\varepsilon }{2}}\). Since \(f\) is Riemann integrable, we have \[U\big (P_1,f\big ) - \frac {\varepsilon }{2}< \overline {\int ^b_a}f(x)\, dx = \underline {\int ^b_a}f(x)\, dx < L\big (P_2,f\big ) + \frac {\varepsilon }{2}\] That is \(\, U\big (P_1,f\big ) - \varepsilon /2 < L\big (P_2,f\big ) + \varepsilon /2\)

Now, considering the common partition of \(P_1\) and \(P_2\) we have \[U\big (P_1\cup P_2,f\big ) - \frac {\varepsilon }{2} < L\big (P_1\cup P_2,f\big ) + \frac {\varepsilon }{2}\] Let \(P_1\cup P_2\) be a single partition \(P\). Then \[U\big (P,f\big ) - \frac {\varepsilon }{2} < L\big (P,f\big ) + \frac {\varepsilon }{2}\] This gives \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \).


□

Proof. Recall that a continuous function on a closed bounded interval is uniformly continuous there.
Suppose that \(f\) is continuous on \([a,b]\) and let \(\varepsilon > 0\) be given, we show that corresponding to this \(\varepsilon > 0\), there is partition \(P\) such that \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \) and then by the above Theorem shall conclude that \(f\) is Riemann integrable on \([a,b]\).
Now by uniform continuity of \(f\) on \([a,b]\), there is a \(\delta > 0\) such that if \(x, \, y\in [a,b]\) with \(\begin {vmatrix} x - y\\ \end {vmatrix} < \delta \) then \(\begin {vmatrix} f(x) - f(y)\\ \end {vmatrix} < \dfrac {\varepsilon }{b - a}\).
Let \(P\) be any partition of \([a,b]\) with \(\begin {vmatrix} \begin {vmatrix} P\\ \end {vmatrix} \end {vmatrix} < \delta \). By the property of continuous functions, on a closed bounded interval \([x_{k -1}, x_k]\), there exists points \(x_k'\) and \(x_k''\) such that \(\, f(x_k') = M_k\,\) and \(\, f(x_k'') = m_k\). Now \begin {align*} \begin {vmatrix} x_k' - x_k''\\ \end {vmatrix} \, & \leq \begin {vmatrix} x_k - x_{k - 1}\\ \end {vmatrix} = \begin {vmatrix} \Delta x_k\\ \end {vmatrix}\\ & < \begin {vmatrix} \begin {vmatrix} P\\ \end {vmatrix} \end {vmatrix}\\ & < \delta \end {align*}

Hence \(\, M_k - m_k = \begin {vmatrix} f(x_k') - f(x''_k)\\ \end {vmatrix} < \dfrac {\varepsilon }{b - a}\,\) for \(\, k = 1,\, 2,\, \cdots \cdots \cdots \, n\). Thus \begin {align*} U\big (P,f\big ) - L\big (P,f\big ) \, & = \sum ^n_{k = 1}\Big [M_k(f) - m_k(f)\Big ]\,\begin {vmatrix} \Delta x_k\\ \end {vmatrix}\\\\ & = \sum ^n_{k = 1}\begin {vmatrix} f(x'_k) - f(x''_k)\\ \end {vmatrix}\, \begin {vmatrix} \Delta x_k\\ \end {vmatrix}\\\\ & < \dfrac {\varepsilon }{b - a}\, \sum ^n_{k = 1}\begin {vmatrix} \Delta x_k\\ \end {vmatrix}\\\\ & = \dfrac {\varepsilon }{b - a}\, (b - a) = \varepsilon \end {align*}

Thus \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \).
Hence from the above theorem, \(f\) is Riemann integrable on \([a,b]\).

□

Proof. If \(f\) is a constant function on \([a,b]\) then it is Riemann integrable on \([a,b]\) by earlier example.
Assume that \(f\) is monotone increasing function on \([a,b]\) and \(f(a) < f(b)\).
Let \(\varepsilon > 0\) be given and let \(P\) be a partition on \([a,b]\) with \(\begin {vmatrix} \begin {vmatrix} P\\ \end {vmatrix} \end {vmatrix} = \dfrac {\varepsilon }{f(b) - f(a)}\). Since \(f\) is increasing on \([a,b]\) we have on \([x_{k - 1}, x_k]\, ,\hspace {0.2cm} M_k(f) =f(x_k)\) and \(m_k(f) = f(x_{k-1})\,\) for
\(\, k = 1,\, 2,\, \cdots \cdots \cdots \, n\). Hence \begin {align*} U\big (P,f\big ) - L\big (P,f\big ) \, & = \sum ^n_{k = 1} M_k(f)\, \begin {vmatrix} \Delta x_k\\ \end {vmatrix}\, -\, \sum ^n_{k = 1} m_k(f)\, \begin {vmatrix} \Delta x_k\\ \end {vmatrix}\\\\ & = \sum ^n_{k = 1} \Big [M_k(f) - m_k(f)\Big ]\, \Delta x_k = \sum ^n_{k = 1} \Big [f(x_k) - f(x_{k - 1})\Big ]\, \Delta x_k\\\\ & < \frac {\varepsilon }{f(b) - f(a)}\, \sum ^n_{k = 1} \Big [f(x_k) - f(x_{k - 1})\Big ]\\\\ & = \dfrac {\varepsilon }{f(b) - f(a)}\, \big (f(x_n)-f(x_0)\big )\\\\ & = \dfrac {\varepsilon }{f(b) - f(a)}\, \big (f(b) - f(a)\big ) = \varepsilon \end {align*}

That is \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \) and so by earlier Theorem, \(f\) is Riemann integrable on \([a,b]\).

\(*\) Prove for the decreasing function.

  □

Proof.

i).

To get the first inequality, \(\, L\big (\sigma _n, f\big ) \leq \dfrac {1}{n}\, \sum \limits ^n_{k = 1} f(x^*_k)\, \), note that \(\, m_k(f)\leq f(x^*_k)\,\) so that \begin {align*} \sum ^n_{k = 1} m_k(f)\, \Delta x_k & \leq \sum ^n_{k = 1}f(x^*_k)\, \Delta x_k\\\\ \implies \hspace {0.4cm} \sum ^n_{k = 1}m_k(f)\, \Delta x_k & \leq \dfrac {1}{n}\, \sum ^n_{k = 1}f(x^*_k)\\\\ \implies \hspace {0.4cm} L\big (\sigma _n, f\big ) & \leq \dfrac {1}{n}\,\sum ^n_{k = 1} f(x^*_k) \end {align*}

The second inequality \(\, \displaystyle {\dfrac {1}{n}\, \sum ^n_{k = 1}f(x^*_k) \leq U\big (\sigma _n , f)}\,\) is obtained in a similar fashion since \(\, f(x^*_k) \leq M_k(f)\,\) for each \(k\).
Hence \(\, \displaystyle {L\big (\sigma _n, f\big ) \leq \frac {1}{n}\, \sum ^n_{k = 1} f(x^*_k) \leq U\big (\sigma _n, f)}\,\) as required.

ii).

Since \(f\) is continuous on \([0,1]\) it is bounded on \([0,1]\), given a \(\varepsilon > 0\) there exits a partition such that \(\, U\big (\sigma _n, f\big ) - L\big (\sigma _n,f\big )< \varepsilon \, \). But \(\, \displaystyle {L\big (\sigma _n,f\big )\leq \int ^1_0f(x)\, dx}\)
\(\displaystyle {\implies \hspace {0.2cm} U\big (\sigma _n,f\big ) - \int ^1_0f(x)\, dx < \varepsilon }\,\) for large \(n\) \[U\big (\sigma _n, f\big ) - L\big (\sigma _n,f\big ) < \varepsilon \] \[U\big (\sigma _n,f\big ) - \int ^1_0f(x)\, dx \, + \, \int ^1_0f(x)\, dx - L\big (\sigma _n,f\big ) < \varepsilon \] But \(\,\, \displaystyle {\int ^1_0f(x)\, dx \, -\, L\big (\sigma _n,f\big ) >0}\). Therefore \(\, \displaystyle {U\big (\sigma _n,f\big ) \, - \, \int ^1_0f(x)\, dx < \varepsilon }\)

Therefore \(\hspace {0.2cm}\displaystyle {\lim _{n\rightarrow \infty }\, U\big (\sigma _n, f\big ) = \int ^1_0 f(x)\, dx}\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (A)\)

Similarly \(\, \, \displaystyle {\int ^1_0f(x)\, dx - L\big (\sigma _n, f\big ) < \varepsilon }\hspace {0.2cm}\) so that \(\hspace {0.2cm} \displaystyle {\int ^1_0f(x)\, dx = \lim _{n\rightarrow \infty } L\big (\sigma _n,f\big )}\hspace {0.3cm}\cdots \cdots \hspace {0.2cm} (B)\)

Combining \((A)\) and \((B)\) gives \[\int ^1_0f(x)\, dx \leq \lim _{n\rightarrow \infty }\, \dfrac {1}{n}\, \sum ^n_{k = 1} f(x^*_k) \leq \int ^1_0f(x)\, dx\]

Hence \(\hspace {0.3cm}\displaystyle {\lim _{n\rightarrow \infty } \, \frac {1}{n}\, \sum ^n_{k = 1}f(x^*_k)=\int ^1_0f(x)\, dx}\)

Proof. Since \(\, \sin x/x \, \) is continuous for all \(x\neq 0\) and \(\, \lim \limits _{x\rightarrow 0} \dfrac {\sin x}{x} = 1 = f(0),\,\) we see that \(f\) is continuous on a closed bounded interval \([0,1]\). Therefore, \(f\) is Riemann integrable on \([0,1]\).


□

Proof.

(i).

If \(f\) is discontinuous on \([a,b]\) then \[M = \sup \big \{f(x):\, x\in [a,b]\big \} \neq \inf \big \{f(x):\, x\in [a,b]\big \} = m\] Let \(\, \big \{\alpha _1,\, \alpha _2,\, \cdots \cdots \cdots ,\, \alpha _p\big \}\,\) be the ordered set of \(p\) points of discontinuity of \(f\) on \([a,b]\). Let \(\alpha _1\neq a\) and \(\alpha _p \neq b\). For \(\varepsilon > 0\), enclose these points of discontinuity in non-overlapping subintervals \(\, [x_1', x''_1]\,,\, [x'_2,x_2'']\, ,\cdots \cdots \cdots ,\, [x_p',x_p'']\,\) of \([a,b]\) such that their total length is less than \(\, \dfrac {\varepsilon }{2\, \big (M - m\big )}\) \[\text {i.e}\hspace {0.5cm}\sum ^p_{k = 1}(x''_k - x'_k)\, < \, \dfrac {\varepsilon }{2\, \big (M - m\big )}\] The function is then continuous on each of the \(p + 1\) subintervals \(\, [a,x_1']\, ,\, [x''_1,x'_2]\, ,\, [x''_2,x'_3]\, , \cdots \, [x''_p,b]\), so for \(\varepsilon > 0\) there exists partitions \(\, P_1,\, P_2,\, \cdots \cdots \cdots ,\, P_{p+1}\,\) of the subinterval such that \[U\big (P_r,f\big ) - L\big (P_r,f\big ) < \dfrac {\varepsilon }{2\,\big (p+1\big )}\hspace {0.4cm} r =1,\, 2,\, \cdots \cdots \cdots ,\, p + 1\] Then for the partition \(\, P= \big \{a = x_0,\, x_1,\, \cdots \cdots \cdots ,\, x_1',\, x_p',\, x_p'',\, \cdots \, x_n = b\big \}\,\) of \([a,b]\) we have \begin {align*} U\big (P,f\big ) - L\big (P,f\big ) & = \sum ^p_{r=1} M_r'(f)\, \big (x''_r-x'_r\big ) + \sum ^{p+ 1}_{r=1}U\big (P_r,f\big ) - \sum ^p_{r=1}m'_r(f)\,\big (x''_r-x'_r\big ) - \sum ^{p + 1}_{r=1}L\big (P_r,f\big )\\\\ & = \sum ^p_{r=1}\big (M'_r(f) - m'_r(f)\big )\,\big (x''_r-x'_r\big ) + \sum ^{p + 1}_{r=1}\Big (U\big (P_r,f\big ) - L\big (P_r,f\big )\Big ) \end {align*}

where \(\hspace {0.2cm}M'_r(f) = \sup \big \{f(x):\, x\in [x'_r,x''_r]\big \}\,\) and \(m'_r(f) = \inf \big \{f(x):\, x\in [x'_r,x''_r]\big \}\,\). Thus \begin {align*} U\big (P,f\big ) - L\big (P,f\big ) & < \sum ^p_{r=1}\big (M -m\big )\, \big (x''_r-x'_r\big ) + \dfrac {\varepsilon }{2\, \big (p + 1\big )}\, \cdot \, \big (p + 1\big )\\\\ & < \big (M-m\big ) \, \dfrac {\varepsilon }{2\big (M -m\big )}\, + \, \frac {\varepsilon }{2} = \varepsilon \end {align*}

Hence \(f\) is Riemann integrable.
The case \(\alpha _1 = a\) or \(\alpha = b\) can easily be dealt with by slight modifications in the above arguments for \(x'_1 = a\) or \(x''_p = b\).

(ii).

To prove the second part, let \(\big \{\alpha _1,\, \alpha _2,\, \cdots \cdots \cdots \, \alpha _p\big \}\) be the ordered set of finite number of limit points of the set of discontinuities of \(f\) on \([a,b]\). Let \(\alpha _1\neq a\) or \(\alpha _p\neq b\). For \(\varepsilon >0\) enclose there are limit points in \(P\) non-overlapping subintervals \([x_1',x_1'']\, ,\, [x'_2,x_2'']\, \cdots \cdots ,\, [x_p',x_p'']\,\) such that their total length is less than \(\, \dfrac {\varepsilon }{2\,\big (M-m\big )}\) \[\text {i.e}\hspace {0.5cm}\sum ^p_{k = 1}\big (x''_k - x'_k\big ) \, < \, \dfrac {\varepsilon }{2\, \big (M-m\big )}\] In each of the remaining \(p + 1\) subintervals \([a,x_1']\, ,\, [x_1'',x'_2]\, , \, \cdots \cdots \cdots \, [x''_p, b]\) of \([a,b]\hspace {0.2cm} f\) has only a finite number of points of discontinuity so that each of these \(p+1\) subintervals can be partitioned to \(P_1,\, P_2,\, \cdots \cdots \cdots \, P_{p+1}\) such that \[U\big (P_r,f\big ) - L\big (P_r,f\big ) < \dfrac {\varepsilon }{2\big (p + 1\big )}\hspace {0.2cm} r = 1,\, 2,\, \cdots \cdots \cdots \, p+1\] Then for the partition \(P\) as in (i) we get \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \).
Hence \(f\) is Riemann integrable.