A convolution semigroup is a collection of probability measures $(\mu_t)_{t \in I}$ on $\mathbb R^d$, where $I \subset [0,\infty)$, for which $\mu_s * \mu_t = \mu_{s+t}$. The convolution semigroup is continuous if $I = [0,\infty)$ and $\delta_0 =\mathop{\textrm{w-lim}}_{t \searrow 0} \mu_t$, where w-lim denotes the weak limit. I want to prove the following:
Assume that $(\nu_t : t \geq 0)$ is a continuous convolution semigroup. Show that $\nu_t = \lim_{s \to t} \nu_s$ for all $t > 0$.
I'm interpreting this limit to mean that the one-sided weak limits $\mathop{\textrm{w-lim}}_{s \searrow t} \nu_s$ and $\mathop{\textrm{w-lim}}_{s \nearrow t} \nu_s$ are both equal to $\nu_t$.
Limit from above: We can calculate directly that if $f$ is a compactly supported continuous function on $\mathbb R^d$, $$\lim_{s\searrow t} \int f \, d\nu_s = \lim_{s \searrow t} \int f\,d(\nu_t * \nu_{s-t}) = \lim_{s \searrow t} \iint f(x+y) \, d\nu_t(x) d\nu_{s-t}(y) = \iint f(x+y) \, d\nu_t(x) d\delta_0(y) = \int f \, d\nu_t$$ by the Portemanteau Theorem, since the function $y \mapsto \int f(x+y) \, d\nu_t(x)$ is continuous and bounded by $\sup f$. Again, by Portemanteau, since $\mathbb R^d$ is Polish and locally compact, and all our measures are probability measures, and since $\mathop{\textrm{v-lim}}_{s \searrow t}\nu_s = \mathop{\textrm{v-lim}}_{s \searrow t} (\nu_{s - t} * \nu_t) = \nu_t$ as we've just shown, we get that $\nu_t = \mathop{\textrm{w-lim}}_{s \searrow t} \nu_s$.
Limit from below: Here I'm stuck. I can show that when $s < t$, we have that $\nu_t = \nu_{t-s} * \nu_s$, and if we knew $\mathop{\textrm{w-lim}}_{s \nearrow t} \nu_s$ exists, we could show $\nu_t = \mathop{\textrm{w-lim}}_{s \nearrow t} \nu_s$ using the same strategy as the previous case.
But how do we know $\mathop{\textrm{w-lim}}_{s \nearrow t} \nu_s$ exists? If the collection $(\nu_s)_{s \geq 0}$ were tight, then we could use Prohorov's Theorem to extract a convergent subsequence of any subsequence of $(\nu_s)$ as $s \to t$. It would follow that $\nu_s$ converges to $\nu_t$. But there's not a clear reason to suspect that $(\nu_s)$ is tight (it's not tight if, for example, $\nu_s = \mathcal{N}_{0, s^2}$ is the Gaussian normal distribution).
So why do we know $(\nu_s)_s$ has a limit point as $s \nearrow t$? Or is there a better way to show $\mathop{\textrm{w-lim}}_{s \nearrow t} \nu_s = \nu_t$?
Both the $s \nearrow t$ and $s \searrow t$ directions can be proven using characteristic functions and Lévy's Continuity Theorem.
The characteristic function $\hat\nu: \mathbb R^d \to \mathbb C$ of a probability measure $\nu$ on $\mathbb R^d$ is given by $$\displaystyle \hat\nu(\xi) = \int_{\mathbb R^d} e^{i\langle \xi, x\rangle} d\nu(x).$$ It's straightforward to show that $\widehat{\nu_t * \nu_s} = \hat\nu_{s+t}$. Let's consider $s \nearrow t$.
Lemma. $\hat\nu_t(\xi) \neq 0$ for any continuous convolution semigroup $(\nu_t)_{t \geq 0}$ on $\mathbb R^d$ and any $\xi \in \mathbb R^d$.
Proof. One can show $|\hat\nu_t|^2$ is the characteristic function of $\nu_t*\tilde\nu_t$ for each $t \geq 0$, where $\tilde\nu_t$ is the probability measure given by $\tilde\nu_t(A) = \nu_t(-A)$. Since $\nu_s$ exists for all $s \geq 0$, we may consider $(\nu_{t/n})_{n \geq 1}$, and observe that $\nu_t = \nu_{t/n}^{*n}$ for all $n \geq 1$. Thus, $\hat\nu_{t/n} = \hat\nu_t^{1/n}$. Therefore, $$ \lim_{n \to \infty}|\hat\nu_{t/n}|^2 = \lim_{n \to \infty} |\hat\nu_t|^{2/n} = \mathbb 1_{\{\hat\nu_t \neq 0\}}. $$ Since $\hat\nu_t(0) = 1$, by continuity of characteristic functions, $\mathbb 1_{\{\hat\nu_t \neq 0\}}$ is continuous at $\xi=0$, so by the Lévy Continuity Theorem, $\mathbb 1_{\{\hat\nu_t \neq 0\}}$ is a characteristic function of a probability measure $\mu$ on $\mathbb R^d$. Again by continuity of characteristic functions, $\mathbb 1_{\{\hat\nu_t \neq 0\}} \equiv 1$. $\tag*{$\Box$}$
We proceed. For $s < t$, observe $\hat\nu_t = \hat\nu_{t-s}\hat\nu_s$. Since $\hat\nu_{t-s}(\xi) \neq 0$ for any $\xi \in \mathbb R^d$, $\hat\nu_t/\hat\nu_{t-s}$ is a continuous and well-defined function. By Lévy's Continuity Theorem and the definition of continuous convolution semigroups, $\hat\nu_t \to 1$ pointwise as $t \to 0$. In particular, $\hat\nu_{t-s} \to 1$ pointwise as $s \nearrow t$. So, $$ \lim_{s \nearrow t} \hat\nu_s = \lim_{s \nearrow t} \frac{\hat \nu_t}{\hat\nu_{t-s}} = \hat\nu_t. $$ By Lévy's Continuity Theorem and uniqueness of characteristic functions, $\nu_s \to \nu_t$ weakly as $s \nearrow t$.
The proof for $s \searrow t$ is similar, but easier.