I am quite lost on this problem. I was hoping someone could give me some direction or hints.
Let $P=\{x_0,x_1,...,x_n\}$ partition $[a,b]$. Define $\|P\|:= \max_{1 \leq i \leq n} \Delta x_i$, the length of the longest interval. Let $f:[a,b] \rightarrow \mathbb{R}$ and $L \in \mathbb{R}$.
We say that $\lim_{\|P\| \rightarrow 0} \sum_{i=1}^nf(t_i)\Delta x_i = L$ if for all $\epsilon > 0$ there exists $\delta > 0$ such that if a partition $P$ satisfies $\|P\|<\delta $ and $t_i \in [x_{i-1}, x_i]$ then $\left| \sum_{i=1}^nf(t_i)\Delta x_i - L \right|<\epsilon $.
Show that $\lim_{\|P\| \rightarrow 0} \sum_{i=1}^nf(t_i)\Delta x_i = L$ if and only if $f$ is Riemann integrable. Also show that $\lim_{\|P\| \rightarrow 0} \sum_{i=1}^nf(t_i)\Delta x_i = L = \int_a^b f(x)dx$.
I know that $L(P,f) \leq \sum_{i=1}^nf(t_i)\Delta x_i \leq U(P,f) \leq \sum_{i=1}^nM_i \delta$. Our professor hinted that we should prove $f$ must be bounded. Any ideas?
Edit:
I figured out one direction, but it seems sketchy. Suppose $f$ is Riemann integrable. Let $\epsilon > 0$ and choose a partition $P$ such that $U(P,f) - L(P,f) < \epsilon$. Recall $L(P,f) \leq \sum_{i=1}^n f(t_i) \Delta x_i \leq U(P,f)$. Then $L(P,f) - L \leq \sum_{i=1}^n f(t_i) \Delta x_i \leq U(P,f) - L$. Recall $U(P,f) - L < \epsilon$ and $L(P,f) - L > -\epsilon$. Then $|\sum_{i=1}^n f(t_i) \Delta x_i| < \epsilon$. $\square$
What I don't like about this proof is it doesn't make use of the size of the partition.
Hint
It's about proving the equivalence of the definitions of Riemann integral a la Darboux and as a limit.
If $f$ is bounded on $[a,b]$, the following statements are valid anyway, that is even if $f$ is not integrable.
$(1)$ $$\lim_{\|P\|\to 0} L(P,f)=\sup_P L(P,f)\qquad\text{and}\qquad\lim_{\|P\|\to 0} U(P,f)=\inf_P U(P,f)$$$(2)$ $$L(P,f)=\inf_T S(P,T,f)\qquad\text{and}\qquad U(P,f)=\sup_T S(P,T,f)$$ where $S(P,T,f)$ is the Riemann sum associated to $P$ tagged by $T$.
Use $(1)$ for the direction you tried, $(2)$ for the other one after proving $f$ is bounded.
You can prove $(1)$ and $(2)$ at a second stage.