Bounded function is Riemann-Integrable if and only if its upper and lower integrals are equal

Question

I was looking at this proof from my measure theory class but specifically im having trouble understanding the $\Leftarrow$ part, that is, proving that for a bounded function $f:[a,b]\rightarrow \mathbb{R}$ if $\mathcal{U}(f)=\mathcal{L}(f)$ then it is Riemann-Integrable, where $\mathcal{U}$ and $\mathcal{L}$ are the upper and lower integrals, respectively. I would appreciate if someone could explain this part of the proof.

Answer 1

Use the following $$\mathcal L(f)=\lim_{\|P\| \to 0} \mathcal L(f,P)$$ and $$\mathcal U(f)=\lim_{\|P\| \to 0} \mathcal U(f,P)$$ which are valid in any case.

Then, if $\varepsilon>0$, there exists $\delta_1>0$ such that $$\mathcal L(f,P)>\mathcal L(f)-\varepsilon$$ if $\|P\|<\delta_1$

and there exists $\delta_2>0$ such that $$\mathcal U(f,P)<\mathcal U(f)+\varepsilon$$ if $\|P\|<\delta_2$ .

Let $\delta=\min\{\delta_1,\delta_2 \}$ .

If $T$ is any set of tags of $P\,$, you have $$\mathcal L(P)-\varepsilon < \mathcal R(f,P,T) < \mathcal U(P)+\varepsilon$$

if $\|P\|<\delta$ because $$\mathcal L(f,P)\le \mathcal R(f,P,T)\le \mathcal U(f,P)$$ So you are done.

Just tell me if you already knew the limits at the start.