Equivalent characterization of Poisson processes

Question

Let $\alpha>0$. We usually say that an $\mathbb N_0$-valued process $(N_t)_{t\ge0}$ on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ is Poisson with intensity $\alpha$ if

1. $N_t-N_s$ is independent of $\mathcal F_s$ for all $t\ge s\ge0$;
2. $N_t\sim\operatorname{Poi}(t\alpha)$ for all $t\ge0$.

Now I've encountered a different definition replacing (2.) by

3. $f(t):=\operatorname E[N_t]<\infty$ for all $t\ge0$; and
4. $f$ is continuous.

It's clear to me that (2.) implies (3.) and (4.). But how can we show the converse?

I'm aware of the following fact, which might be helpful to show the desired implication: If $g:[0,\infty)\to\mathbb C$ is right-continuous and $g(s+t)=g(s)g(t)$, then $g(t)=g(1)^t$ for all $t\ge0$.