Prove that a bounded function $f$ is integrable on $[0,1]$ if $$I_0 := \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n),$$ in which case $\int_0^1f(x)dx$ equals $I_0$.
Refer here. I suspect that this is the same question with answers but I am not sure how to apply it to prove my particular case.
Furthermore, consider the definition that I was given below
Definition: $f : [a,b]\to \mathbb R$ is said to be (Riemann) integrable on $[a,b]$ if and only if $f$ is bounded on $[a,b]$, and for every $\epsilon > 0$ there is a partition $P$ of $[a,b]$ such that $U(f,P) - L(f,P) < \epsilon$.
With this definition in mind, why then did the linked post above need to show that $$\overline{\int}_a^b f \leq U(f,P_N) \leq L(f,P_N) \leq \underline{\int}_a^b f?$$ Would it not just follows from the definition (in my particular case) that if $U=L$ as $n\to \infty$ then there automatically is a partition $P$ such that $U(f,P) - L(f,P) < \epsilon$ is satisfied by definition?
Any clarification is helpful thank you!
Beccause the upper and lower integrals are the infimum and supremum, respectively, of upper and lower sums, it follows that for any $N$ and $M$
$$L(f,P_N) \leqslant \underline{\int}_a^b f \leqslant \overline{\int_a}^bf\leqslant U(f,P_M)$$
Since $I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$, for any $\epsilon > 0$ there exists $N$ and $M$ such that
$$I_0 - \epsilon < L(f,P_N) \leqslant U(f, P_M ) < I_0 +\epsilon$$
Thus,
$$I_0 - \epsilon < \underline{\int}_a^b f \leqslant \overline{\int_a}^bf< I_0 +\epsilon$$
Since $\epsilon$ can be arbitrarily close to $0$, it follows that
$$I_0= \underline{\int}_a^b f = \overline{\int_a}^bf$$
This proves both that $f$ is Riemann integrable and that $I_0$ is the value of the integral.