Rigorous Proof?: Proving Cauchy Criterion of Integrals I am trying to prove the following: (I only know about Riemann Integrals)

Let $f$ be a bounded function on an interval [a, b] and assume that $P_n$
  is a sequence of partitions of [a, b] such that $\lim\limits_{n\to\infty} L(f, P_n) = \lim\limits_{n\to\infty} U(f, P_n) = s.$
Then, $f$ is integrable and $\int_a^bf dx = s.$

Proof:
We know that if $f$ is bounded on $[a,b]$, then $f$ is integrable on [a,b] if and only if for every $\varepsilon \gt0$ there is a partition $P$ of $[a,b]$ such that $U(f,P)−L(f,P)\lt\varepsilon.$
From our assumption, we know that there exists a partition $P_n$ s.t. $U(f,P_n)−L(f,P_n)=s-s=0\lt\varepsilon$ when $n\to\infty$
Therefore, $f$ is integrable.
Since $f$ is integrable, $L(f,P)=U(f,P)= \int_a^bf dx$,
$\int_a^bf dx=s$
Question (Comment):
I feel like what I have is very "weak" and unclear. I'm also unsure whether or not the last part even makes sense. Thank you.
Notation: U(f,P) indicates the Riemann Upper Sum and L(f,P), the lower sum.

For every $\varepsilon\gt0$, one wants to exhibit a partition $P$ such that $U(f,P)-L(f,P)\lt\varepsilon$. 
Since $U(f,P_n)\to s$ when $n\to\infty$, there exists $n_U(\varepsilon)$ such that for every $n\geqslant n_U(\varepsilon)$, $|U(f,P_n)-s|\lt\frac12\varepsilon$. Since $L(f,P_n)\to s$ when $n\to\infty$, there exists $n_L(\varepsilon)$ such that for every $n\geqslant n_L(\varepsilon)$, $|L(f,P_n)-s|\lt\frac12\varepsilon$. 
Choose $n^*(\varepsilon)=\max\{n_U(\varepsilon),n_L(\varepsilon)\}$ and $P=P_{n^*(\varepsilon)}$. Then 
$$
U(f,P)-L(f,P)\leqslant
|U(f,P_{n^*(\varepsilon)})-s|+|L(f,P_{n^*(\varepsilon)})-s|\lt
\tfrac12\varepsilon+\tfrac12\varepsilon=\varepsilon.
$$
This proves that $f$ is integrable.
Furthermore, $L(f,P_n)\leqslant \int\limits_a^b f\leqslant U(f,P_n)$ for every $n$, hence 
$$
\sup\limits_n\ L(f,P_n)\leqslant \int\limits_a^b f\leqslant \inf\limits_n\ U(f,P_n).
$$
The supremum on the LHS and the infimum on the RHS are both equal to $s$ hence $\int\limits_a^b  f=s$.
