Proof of Continuous Functions on a Closed Interval are Riemann Integrable

Let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$. We must show that $f$ is Riemann integrable on $[a,b]$.

By Continuity on a Closed Interval Implies Uniform Continuity, the function $f$ is uniformly continuous on $[a,b]$. Fix $\varepsilon>0$. Choose $\delta>0$ such that for all $s,t\in[a,b]$, if $|s-t|<\delta$, then

$$|f(s)-f(t)|<\varepsilon.$$

For each positive integer $n$, let $P_n$ be the partition of $[a,b]$ into $2^n$ equal subintervals, so that the mesh of $P_n$ is $(b-a)/2^n$. Choose $N$ so large that $(b-a)/2^N<\delta$. For each $n\ge N$, choose a tagged partition $\mathcal{T}_n$ whose underlying partition is $P_n$, and let

$$R_n = R(f,\mathcal{T}_n)$$

be its Riemann sum.

We claim that $(R_n)_{n=N}^\infty$ is a Cauchy sequence. Let $m\ge n\ge N$. Since $P_m$ refines $P_n$, each subinterval of $P_m$ lies inside a unique subinterval of $P_n$. Construct a tagged partition $\mathcal{S}_{m,n}$ on the partition $P_m$ by assigning to each subinterval of $P_m$ the tag of the containing subinterval of $P_n$. Then

$$R(f,\mathcal{S}_{m,n})=R_n,$$

because subdividing a coarse interval without changing its tag does not change its total contribution to the sum.

Now $\mathcal{S}_{m,n}$ and $\mathcal{T}_m$ have the same underlying partition $P_m$. On each subinterval of $P_m$, both associated tags lie in that subinterval, whose length is at most the mesh of $P_m$, hence less than $\delta$. Therefore the choice of $\delta$ gives

$$|f(\sigma)-f(\tau)|<\varepsilon$$

for the two tags on each such subinterval. Summing over the subintervals of $P_m$, we obtain

$$|R_n-R_m| = |R(f,\mathcal{S}_{m,n})-R(f,\mathcal{T}_m)| \le \varepsilon \sum_{i=1}^{2^m}(x_i-x_{i-1}) = \varepsilon(b-a).$$

Since the original positive number was arbitrary, we may begin instead with $\varepsilon/(b-a)$; hence for every $\varepsilon>0$ there exists $N$ such that $|R_n-R_m|<\varepsilon$ whenever $m,n\ge N$. Thus $(R_n)$ is a Cauchy sequence of real numbers.

By Every Cauchy Sequence of Real Numbers Converges, there exists $I\in\mathbb{R}$ such that $R_n\to I$ in the sense of Limit of a Sequence of Real Numbers. We now show that this number $I$ is the Riemann integral of $f$.

Let $\eta>0$. Apply uniform continuity with $\eta/(2(b-a))$ in place of $\varepsilon$, and choose $N_1\ge N$ so large that $(b-a)/2^{N_1}<\delta$ and $|R_n-I|<\eta/2$ for all $n\ge N_1$. Let $\mathcal{U}$ be any tagged partition of $[a,b]$ whose mesh is less than $\delta$, and choose $n\ge N_1$. Let $Q$ be the common refinement of the underlying partition of $\mathcal{U}$ and the dyadic partition $P_n$. Construct tagged partitions $\mathcal{U}'$ and $\mathcal{T}'_n$ on $Q$ by retaining on each refined subinterval the tag coming from the containing subinterval of $\mathcal{U}$ or $\mathcal{T}_n$, respectively. Then

$$R(f,\mathcal{U}')=R(f,\mathcal{U}),\qquad R(f,\mathcal{T}'_n)=R_n.$$

Moreover, on each subinterval of $Q$, the two retained tags lie in the same subinterval of $Q$, whose length is less than $\delta$. Therefore

$$|R(f,\mathcal{U})-R_n| = |R(f,\mathcal{U}')-R(f,\mathcal{T}'_n)| < \frac{\eta}{2(b-a)}(b-a)=\frac{\eta}{2}.$$

Hence

$$|R(f,\mathcal{U})-I|\le |R(f,\mathcal{U})-R_n|+|R_n-I|<\eta.$$

Thus every tagged partition of $[a,b]$ with sufficiently small mesh has Riemann sum within $\eta$ of $I$. By Riemann Integrability on a Closed Interval, the function $f$ is Riemann integrable on $[a,b]$, and

$$\int_a^b f(t)\,dt = I.$$