let $f$ be continuous on $[a,b]$. Prove that $f$ is integrable on $[a,b]$

Question

I know I will need to use the definition of uniform continuity but I don't know how to go about proving $f$ is integrable

Answer 1

We will use the following result.

Let $a,b\in\mathbb R,a<b$ and $f:[a,b]\rightarrow\mathbb R$. The following are equivalent:

1. $f$ is Riemann-integrable
2. For $\varepsilon>0$ there exist step functions $\varphi,\psi$ with $\varphi\leq f\leq \psi$ and $$\int\limits_{a}^{b}\! \psi(x)-\varphi(x)\,\mathrm{d}x\leq \varepsilon.$$

Outline of the proof for: $f[a,b]\rightarrow$ is continuous, then $f$ is integrable.

As $[a,b]$ is a compact intervall, there exists $\delta>0$ so that for all $x,y\in[a,b]$ with $|x-y|<\delta$: $$|f(x)-f(y)|<\frac{\varepsilon}{2(b-a)}.$$

Choose $n\in\mathbb N$ with $\dfrac{b-a}{n}<\delta$. Define a partiton $(x_i)_{0\leq i\leq n}$ of the intervall $[a,b]$ via $$x_i:=a+i\cdot\frac{b-a}{n},\quad 0\leq i\leq n.$$ Now let $$\varphi(x):=f(x_i)-\frac{\varepsilon}{2(b-a)},\psi(x):=f(x_i)+\frac{\varepsilon}{2(b-a)},~\quad x_{i-1}\leq x<x_i,1\leq i\leq n.$$

Now we just have to put everything together (which is left as an excrcise to you).

Answer 2

As you have said, we do need to use the definition of uniform continuity. Given $\epsilon > 0 \; \exists\delta>0, \; \text{with } \delta<b-a, \; \text{such that given }x,y\in[a,b],$ $$|x-y|<\delta\implies|f(x)-f(y)|<\frac{\epsilon}{b-a}$$

Now let $\{x_0,x_1,...x_n\}$ be a partition of $[a,b]$ such that the maximum distance between $x_i$ and $x_{i+1}$ is $\delta$. Now continuous functions on closed intervals obtain its supremum and infimum on the interval, so there exist $u_k,v_k\in[x_k,x_{k+1}]$ such that $$f(u_k) = \text{sup}\{f(x):x\in[x_k,x_{k+1}]\}$$ $$f(v_k) = \text{inf}\{f(x):x\in[x_k,x_{k+1}]\}$$ By our assumption that $x_{k+1} - x_k<\delta$, we have $|u_k-v_k|<\delta$. So by our definition of uniform continuity, we have that $$f(u_k)-f(v_k)<\frac{\epsilon}{b-a}$$ Then we have that the difference between the upper sum and the lower sum is: $$\sum_{k=0}^{n-1} f(u_k)(x_{k+1}-x_k)-\sum_{k=0}^{n-1} f(v_k)(x_{k+1}-x_k) \\=\sum_{k=0}^{n-1} [f(u_k)-f(v_k](x_{k+1}-x_k)\\<\frac{\epsilon}{b-a}\sum_{k=0}^{n-1} (x_{k+1}-x_k)\\=\frac{\epsilon}{b-a}(x_n-x_0) = \epsilon$$