Smith's Key Renewal Theorem for Renewal Function

Question

Consider a renewal process $(N_t)_{t \geq 0}$ and its renewal function $M(t):=\mathbb{E}[N_t]$ with interarrival distribution function $F$.

One can show that $M$ satisfies the $(F,F)$-renewal equation, i.e. satisifies

$$M(t)=F(t) + M* F(t) = F(t)+ \int_{0}^{t} M(t-s) d F(s).$$

Smith's key renewal theorem states that for $g$ a solution to the $(h,F)$-renewal equation satisfies

$$\lim_{t \rightarrow \infty} g(t)= \frac{1}{\mathbb{E}[T_i]} \int_0^{\infty} h(s) ds,$$ where $T_i$ is distributed according to $F$, $F$ is non-arithmetic and $h$ is directly Riemann integrable.

So I wonder whether this gives a stronger result than maybe the elementary renewal theorem ($M(t)/t \rightarrow 1/\mathbb{E}[T_i]$) or Blackwell's renewal theorem $(M(t+h)-M(t)\rightarrow h/\mathbb{E}[T_i]$) for $M$ as $t \rightarrow \infty$.

Plugging in yields

$$M(t) \rightarrow \frac{1}{\mathbb{E}[T_i]} \int_0^{\infty} F(s) ds.$$

Now I am not sure whether one can simplify this result. Is it stronger, weaker, different from the others? I have a "basic" lecture on this topic but we did not talk about $M(t)$ in the context of this asymptotics, so I really wonder why we have omitted it. Is $F$ directly riemann integrable at all?

Answer 1

The key renewal theorem and Blackwell's renewal theorem are equivalent; the elementary renewal theorem is implied by but is not equivalent to the other two. This is spelled out in Feller, An Introduction to Probability Theory and its Applications, vol. 2.

[EDIT] The integral of F(s)ds from 0 to infinity is not correct. The value of the integral is infinity. F is not directly (or otherwise) integrable over (0,infinity).