Suppose $\{N(t):t\geqslant0\}$ is a renewal process with renewal times $T_n$, i.e. $N(t) = \sum_{n=1}^\infty \mathsf 1_{[0,t]}(T_n)$ and finite mean $\mu = \mathbb E[T_1]$. Show that $$\mathbb E\left[\int_0^{T_n}N(s)\ \mathsf ds\right] = \mu n(n-1)/2. $$
For a non-random $t>0$ it follows by Fubini's theorem that $$ \mathbb E\left[ \int_0^t N(s)\ \mathsf ds\right] = \int_0^t \mathbb E[N(s)]\ \mathsf ds $$ since $\mathbb P(N(s)=\infty)=0$ for all $s$, hence $s\mapsto N(s)$ is bounded on the compact domain $[0,t]$ so that $$\iint_{\Omega\times[0,t]}N(s)(\omega)\ \mathsf d(s \times \mathbb P)<\infty. $$ I do not see how to show this claim for an arbitrary renewal time $T_n$, however. Would induction be a good approach?
We can directly evaluate $$\begin{align*}\int_0^{T_n} N(s)\,ds &= \sum_{k=0}^{n-1} \int_{T_k}^{T_{k+1}} N(s)\,ds \\ &= \sum_{k=0}^{n-1}\int_{T_k}^{T_{k+1}} k\,ds \\ &= \sum_{k=0}^{n-1}k(T_{k+1}-T_k)\end{align*}$$
Now take the expectation and use linearity: $$\begin{align*}\mathbb{E}\left[\int_0^{T_n} N(s)\,ds\right] &= \mathbb{E}\left[\sum_{k=0}^{n-1}k(T_{k+1}-T_k)\right] \\ &= \sum_{k=0}^{n-1}k\mathbb{E}[T_{k+1}-T_k] \\ &= \sum_{k=0}^{n-1}k\mathbb{E}[T_1] \\ &= \mu n(n-1)/2\end{align*}$$