1

Riemann integral of continuous nonnegative \(f: \mathbb{R}\to\mathbb{R}\) agrees with Lebesgue integral.

Following hint, without loss of generality assume \([a,b]=[0,1]\).

Define \[\begin{align*} a_k^{(n)} &= \min\left\{f(x)| x\in\left[\frac{k-1}n, \frac{k}n\right]\right\} \\ b_k^{(n)} &= \max\left\{f(x)| x\in\left[\frac{k-1}n, \frac{k}n\right]\right\} \\ \underline{f}_n(x) &= \begin{cases} a_k^{(n)} &\text{if } x\in \left[\frac{k-1}n,\frac{k}n\right) \\ a_n^{(n)} &\text{if } x\in \left[\frac{n-1}n,1\right] \\ \end{cases} \\ \overline{f}_n(x) &= \begin{cases} b_k^{(n)} &\text{if } x\in \left[\frac{k-1}n,\frac{k}n\right) \\ b_n^{(n)} &\text{if } x\in \left[\frac{n-1}n,1\right] \\ \end{cases} \\ \underline{I}_n &= \frac1n \sum_{n=1}^N a_k \\ \overline{I}_n &= \frac1n \sum_{n=1}^N b_k \\ \end{align*}\]

For any \(n\), we can see that \[\underline{f}_n(x) \leq f(x) \leq \overline{f}_n(x),\] and therefore \[\underline{I}_n = \int_{[0,1]} \underline{f}_n\ dx \leq \int_{[0,1]} f\ dx \leq \int_{[0,1]}\overline{f}_n\ dx = \overline{I}_n.\]

Let \(\epsilon>0\) be given, as \(f\) is uniformly continuous on the unit interval choose \(N\) large enough such that for all \(\left\lvert x-y\right\rvert<\frac1N\) implies that \(\left\lvert f(x)-f(y)\right\rvert<\epsilon\), then for \(n>N\), \[\overline{I}_n-\underline{I}_n = \int_{[0,1]} \overline{f}_n - \underline{f}_n\ dx < \epsilon.\] This shows that both \(\overline{I}_n\) and \(\underline{I}_n\) converge to the Riemann integral of \(f\).

Now any step function \(s\leq f\) satisfies \(\int_{[a,b]} s\ dx = \int_{[a,b]} s\ d\lambda \leq \overline{I}_n\) for all \(n\), which shows that its Riemann integral coincides with Lebesgue integral.

2

We use definition of Riemann integral from Rudin’s Principles of Mathematical Analysis, so a bounded \(f:[a,b]\to\mathbb{R}\) is integrable if the infimum of all upper sums equal the supremum of all lower sums.

For any partition \(P\) of \([a,b]\) (Rudin did not use tags) we can define the upper and lower step functions \[\begin{align*} \overline{f}_P(x) = f(a)\chi_{\left\{a\right\}} + \sum_i M_i \chi_{(x_i, x_{i+1}]} \\ \underline{f}_P(x) = f(a)\chi_{\left\{a\right\}} + \sum_i m_i \chi_{(x_i, x_{i+1}]} \end{align*}\] where \(M_i = \sup_{x\in[x_i,x_{i+1}]} f(x)\) and \(m_i = \inf{x\in[x_i,x_{i+1}]} f(x)\).

It then arises from the definitions that \[\begin{align*} U(P,f) &= \int_a^b \overline{f}_P(x)\ dx \\ L(P,f) &= \int_a^b \underline{f}_P(x)\ dx \end{align*}\] and as \(f\) is Riemann integrable we have \(\sup_P L(P,f) = \inf_P U(P,f) = \int_a^b f\ dx\).

Let \(P_n\) be a sequence of partitions of \([a,b]\) such that each \(P_{k+1}\) is a refinement of \(P_k\) and the mesh goes to \(0\), and it witnesses the integrability of \(f\), that is \[ \lim_n L(P_n,f) = \int_a^b f\ dx = \lim_n U(P_n,f). \] Then we have for all \(x\in[a,b]\), \[ \underline{f}_{P_1}(x)\leq \underline{f}_{P_2}(x)\leq \dots \leq f(x) \leq \dots \leq \overline{f}_{P_2}(x) \leq \overline{f}_{P_1}(x)\] by monotone convergence \[\int \lim_n \underline{f}_{P_n}\ d\lambda = \lim_n \int\underline{f}_{P_n}\ d\lambda = \int_a^b f\ dx \] and similarly for \(\overline{f}_{P_n}\). This implies that \[ \lim_n \underline{f}_{P_n} = f = \lim_n \overline{f}_{P_n} \quad \text{a.e.} \] which shows that \(f\) is Lebesgue measurable and its Lebesgue integral coincides with the Riemann integral.