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Let $f:[a,b]\to \mathbb{R}$ be a bounded function, with $a<b$. A partition of an interval is $\mathcal{P}=(a=x_0<x_1<\cdots <x_N=b)$; its mesh of a partition is $\Vert \mathcal{P}\Vert ={\max }_i(x_i-x_{i-1})$. Choose arbitrary tags ${\xi }_i\in [x_{i-1},x_i]$. The corresponding Riemann sum isThe Riemann sum convergence theorem says that $f$ is Riemann integrable with Riemann integral $I$ if and only if, for every $\epsilon >0$, some $\delta >0$ satisfiesfor every partition of an interval with $\Vert \mathcal{P}\Vert <\delta$ and every permitted choice of tags. Equivalently, along every sequence of such tagged partitions of an interval with mesh of a partition tending to zero, the Riemann sums tend to $I$. Both the shrinking mesh of a partition and freedom of tags are essential.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

For $a<b$, choose the uniform partition of an interval into $n$ pieces and its midpoint tagsTheir arithmetic mean is exactly $(a+b)/2$. Apply the given Jensen inequality with equal weights ${\lambda }_i=1/n$:Multiplying by $b-a$ gives a midpoint Riemann sum on the left. Its mesh of a partition is $(b-a)/n\to 0$, so the Riemann sum convergence theorem givesFor $a=b$ the equality is immediate. Together with part i, this is the Hermite-Hadamard inequality with the directions appropriate to a concave function.

For a bounded function on a fixed closed interval, Riemann integrability with Riemann integral $I$ is equivalent to convergence of all tagged Riemann sums to $I$ as the mesh of a partition tends to zero. Precisely, for every $\epsilon >0$ there is $\delta >0$ such that every partition of an interval of mesh less than $\delta$, with any tags in the corresponding closed subintervals, has its Riemann sum within $\epsilon$ of $I$.