I want to show that the composite trapezoidal rule for any $f\in C^1[a,b]$:
$T_M=\displaystyle\frac{h}{2}\sum_{k=1}^M[f(x_{k-1})+f(x_k)]$ with $h=(b-a)/M$ and $x_k=a+kh$ for $k=1,2,\ldots,M$
converges to $\int_a^bf(x)dx$.
I know I can use error analysis to show that this is true if $f\in C^2[a,b]$, but I'm running into a "brain block" while trying to show this for simply continuous functions.
Let $\omega_f$ be the modulus of continuity of $f$ on $[a,b]$. On each subinterval $[x_{k-1},x_k]$ we have $|f(x)-f(x_k)|\le \omega_f(h)$. Hence, $$ \left|\int_{x_{k-1}}^{x_k} f(x)\,dx - h f(x_k)\right|\le h\omega_f(h) $$ Since also $|\frac12(f(x_{k-1})+f(x_k))-f(x_k)|\le \frac12 \omega_f(h)$, it follows that $$ \left|\int_{x_{k-1}}^{x_k} f(x)\,dx - \frac{h}{2} f(x_{k-1})+f(x_k)\right|\le \frac32h\omega_f(h) $$ Sum over $k$ to obtain $$ \left|\int_a^b f(x)\,dx - T_M \right|\le \frac32(b-a)\omega_f(h) $$ which converges to $0$ as $h\to 0$.
Note that in contrast to the $C^2$ case (where the error is $O(h^2)$), here we have no information about the rate of convergence: we can't even guarantee that it's $O(h^\alpha)$ for some $\alpha>0$.