Let $\mu_n$ be the measure defined by $$\mu_n(A) = \frac{1}{\tau_n} \int_0^{\tau_n} \mu(A-t)\,dt.$$
Verify that $\mu_n$ is indeed a measure (i.e. that it is countably additive). Use the monotone convergence theorem.
For a bounded measurable function $f : \mathbb{R} \to \mathbb{R}$, verify that $$\int_{\mathbb{R}} f\,d\mu_n = \frac{1}{\tau_n} \int_0^{\tau_n}\int_{\mathbb{R}} f(x+t)\,\mu(dx)\,dt.$$ Start with the case where $f=1_A$ is an indicator function, then do simple functions, then nonnegative measurable functions, then bounded measurable functions (the "standard mantra").
Use Fubini's theorem to verify that $$\tag{*} \int_{\mathbb{R}} f\,d\mu_n = \int_{\mathbb{R}} \frac{1}{\tau_n} \int_0^{\tau_n}f(x+t)\,dt\,\mu(dx).$$
For bounded continuous $f$, set $f_n(x) = \frac{1}{\tau_n} \int_0^{\tau_n}f(x+t)\,dt$. Use the continuity of $f$ to show that $f_n \to f$ pointwise, and the sequence is bounded by $\|f\|_\infty$.
Use the dominated convergence theorem on (*) to show that $\int_\mathbb{R} f\,d\mu_n = \int_{\mathbb{R}} f_n\,d\mu \to \int_\mathbb{R} f\,d\mu$. Hence $\mu_n \to \mu$ weakly.