Prove that if $f \in C[a,b]$, and $T_n(f)$ is the composite trapezoidal approximation of $\int_{a}^{b}f(x)dx$, then $\lim_{n\to\infty}T_n=\int_{a}^{b}f(x)dx$.
$T_n(f)=\frac{h}{2}\sum_{n=1}^{M}(f(x_{n-1})+f(x_n))$ at least I think this is what I am trying to get at.
I do believe I need to use Riemann as well, but I am not sure how to do it.
Since $f$ is continuous, it is Riemann integrable with integral $I = \int_a^b f(x) \, dx.$
We can use a more general non-uniform partition $P: a = x_0 < x_1 < \ldots < x_n = b$ to prove this.
Using the triangle inequality
$$\left|\frac1{2}\sum_{i=0}^{n-1}(\,f(x_i)+f(x_{i+1})\,)(x_{i+1}-x_i)-I \right| \\ = \left|\frac1{2}\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_i)-\frac1{2}I+\frac1{2}\sum_{i=0}^{n-1}f(x_{i+1})(x_{i+1}-x_i)-\frac1{2}I\right|\\\leqslant \frac1{2}\left|\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_i)-I\right|+\frac1{2}\left|\sum_{i=0}^{n-1}f(x_{i+1})(x_{i+1}-x_i)-I\right|.$$
Since $f$ is Riemann integrable, given $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ with mesh $\|P\| = \max_{1\leqslant i \leqslant n}(x_{i}-x_{i-1}) $ less than $\delta$, each absolute value on the RHS is less than $\epsilon$.
Thus, $||P|| < \delta$ implies
$$\left|\frac1{2}\sum_{i=0}^{n-1}(\,f(x_i)+f(x_{i+1})\,)(x_{i+1}-x_i)-I\right| < \epsilon.$$
For a uniform partition, where $x_i = a + \frac{i}{n} (b-a)= a+ih$ and $\|P\| = \frac{b-a}{n}$, we have for all $n > \frac{b-a}{\delta}$,
$$\left|\frac{b-a}{2n}\sum_{i=0}^{n-1}[f(x_i)+f(x_{i+1})]-I\right| < \epsilon,$$
Thus,
$$\lim_{n \to \infty} \frac{h}{2}\sum_{i=0}^{n-1}(\,f(x_i)+f(x_{i+1})\,)=\lim_{n \to \infty} \frac{b-a}{2n}\sum_{i=0}^{n-1}(\,f(x_i)+f(x_{i+1})\,)=I = \int_a^b f(x) \, dx$$
Aside
Proving, for a bounded function, that integrability in the trapezoidal sense implies Riemann integrability is more difficult, but that is another question.
