Sequential characterization of integrability

Question

I'm having some difficulties trying to understand a step from the following proof, which is about the sequential characterization of integrability. Here's the theorem:

And here's the part of the proof I cannot totally understand:

I understand why squeeze theorem can be applied, and I agree with the fact that it implies:

$$ \lim_{n\rightarrow \infty}{[U(f;P_n)-U(f)]} = 0 $$

Yet still I don't see why the above limit can be 'split' into two different limits, since, in general, the fact that: $$ \lim_{n\rightarrow \infty}{(a_n-b_n)} = 0 $$ does not imply the existence of the following limits: $$ \lim_{n\rightarrow \infty}{a_n}, \,\,\,\,\,\,\,\, \lim_{n\rightarrow \infty}{b_n}. $$

Thanks beforehand!!

Answer 1

Since $U(f) = L(f) = \int_a^b f(x) \, dx$, we have

$$L(P_n,f) \leqslant \int_a^bf(x) \, dx \leqslant U(P_n,f),$$

and

$$0 \leqslant U(P_n,f) - \int_a^bf(x) \, dx \leqslant U(P_n,f) - L(P_n,f)\underset{n \to \infty}\longrightarrow 0,$$

which implies $U(P_n,f) \to \int_a^b f(x) \,dx$ as $n \to \infty$.

We can show in a simlar way that $L(P_n,f) \to \int_a^b f(x) \,dx$.

Answer 2

In your case you should denote $a_n:=L(f;P_n)$, $c:=L(f)$, $d:=U(f)$ and $b_n:=U(f;P_n)$, where

$$ a_n \leq c\leq d\leq b_n. $$

You know that $a_n-b_n\to 0$, which as you said implies that

$$ a_n-d\to0 \quad \text{and} \quad b_n-c\to 0. $$