Let $\{X_n:n=1,2,\ldots\}$ be a sequence of i.i.d. random variables with finite mean $\mu$ and assume there exists no $d\geqslant0$ such that $\sum_{n=0}^\infty \mathbb P(X_1=nd)=1$ (i.e. the renewal process is non-lattice). Define $S_n = \sum_{i=1}^n x_i$ for positive integers $n$ and $N(t) = \sup\{n: S_n\geqslant t\}$ for $t\geqslant 0$. Let $m(t) = \mathbb E[N(t)]$ be the renewal function.
Blackwell's Theorem states that for all $a\geqslant 0$, $$ \lim_{t\to\infty} m(t+a)-m(t) = \frac a\mu. $$
Let $h:[0,\infty]\to\mathbb R$ be a directly Riemann integrable function, that is $$ \lim_{a\to 0}\ \sum_{n=1}^\infty \sup\{h(t) : (n-1)a\leqslant t\leqslant na\} = \lim_{a\to 0}\ \sum_{n=1}^\infty \inf\{h(t) : (n-1)a\leqslant t\leqslant na\} $$
The Key Renewal Theorem states that $$ \lim_{t\to\infty} \int_0^t h(t-x)\ \mathsf dm(x) = \frac1\mu \int_0^\infty h(t)\ \mathsf dt.\tag1 $$
From Blackwell's Theorem we see that $$ \lim_{a\to 0} \lim_{t\to\infty} \frac{m(t+a)-m(t)}a = \frac1\mu.\tag2 $$ Ross's Stochastic Processes (p. 112) says:
Now assuming that we can justify interchanging the limits, we obtain $$ \lim_{t\to\infty} \frac{\mathsf dm(t)}{\mathsf dt}= \frac1\mu.\tag3 $$ The key renewal theorem is a formalization of the above.
I am not sure how to justify the interchange of limits in $(2)$, but it is clear that, assuming $m$ were differentiable, $(3)$ would be the result. However, it is not at all obvious that $m$ should be differentiable, aside from when the $X_i$ are absolutely continuous. See for example this question: Convolution form of the renewal density
I also do not see how one arrives at $(1)$ from $(3)$. Here it seems we must interpret e.g. $\mathsf dm(x)$ in the Riemann-Stieltjes sense, but it is not clear where the relation between the (limiting) convolution of $h$ with $\mathsf dm$ and the integral of $h$ over $[0,\infty)$ comes from. How can we show this rigorously?
