A trapezoidal approximation for a sequence of uniformly converging functions Let $f_n : [0,1] \rightarrow \mathbb R$ be a sequence of continuous functions converging uniformly to a function $f$. For each $N \in \mathbb N$, also define a sequence of points $0=x_1< x_2 <...<x_N=1$.
Does it hold that:
$$ \lim_{N \rightarrow\infty} \sum_{k=1}^{N} \frac{f_N(x_{k-1}) + f_N(x_k)}{2} (x_k - x_{k-1}) = \int_0^1 f(x) dx$$
I have an intuition that this should hold, but I have no idea how to prove it.
 A: Since $f_n(x) \to f(x)$ uniformly on $[0,1]$ with $f_n$ continuous, it follows that $f$ is integrable.  As long as the partition $P_N = (x_0,x_1, \ldots, x_N)$ is constructed for each $N$ in such a way that the norm $\|P_N\| \to 0$ as $N \to \infty$, then any tagged Riemann sum $S(P_N,f,T)$ converges to the integral regardless of the choice of tags. 
In particular, this convergence is true for left- and right-hand sums as well as the average of the two: 
$$\tag{1}\lim_{N \to \infty}A(P_N,f) = \lim_{N \to \infty} \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}) = \int_0^1 f(x) \, dx.$$
Note that 
$$\tag{2}A(P_N,f_N) = \sum_{k=1}^N \frac{1}{2}(f_N(x_{k-1}) + f_N(x_{k}))(x_k - x_{k-1}) \\= \sum_{k=1}^N \frac{1}{2}\{[f_N(x_{k-1})-f(x_{k-1})] + [f_N(x_{k})-f(x_k)]\}\,(x_k - x_{k-1}) + \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}). $$
Hence, using (2) and applying the triangle inequality we have
$$\left|A(P_N,f_N) - \int_0^1 f(x) \, dx\right| \\ \leqslant \frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) +\left|A(P_N,f) - \int_0^1 f(x) \, dx\right|.$$
By uniform convergence, for any $\epsilon > 0$ there exists $N_1$ such that if $N > N_1$, then $|f_N(x) - f(x)| < \epsilon/2 $ for every $x \in [0,1]$ and 
$$\frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) < \epsilon/2,$$
since $\sum_{k=1}^N(x_k - x_{k-1}) = 1$.
By the convergence in (1) there exists $N_2$ such that if $N > N_2$, then $$\left|A(P_N,f) - \int_0^1f(x) \, dx\right| < \epsilon/2.$$
Therefore, if $N > \max(N_1,N_2)$ we have $|A(P_N,f_N) - \int_0^1 f(x)\, dx | < \epsilon, $  implying that 
$$\lim_{N \to \infty}A(P_N, f_N) = \int_0^1 f(x) \, dx.$$
A: Hint: Define, for any $g$ on $[0,1],$
$$S_N(g) = \sum_{k=1}^{N}\frac{g((k-1)/N) + g(k/N)}{2}\frac{1}{N.}$$
Then $S_N(g)| \le \|g\|_\infty.$ Furthermore, if $g$ is Riemann integrable on $[0,1],$ then $S_N(g) \to \int_0^1 g.$
So consider in your problem $S_N(f_N) - S_N(f) = S_N(f_N-f).$
