Consider the metric space $X = \mathbb{R}$, $\mathcal{B}$ the Borel $\sigma$-algebra on $\mathbb{R}$ and $\mu$ a probability measure on $X$. Let $A \in \mathcal{B}$ and $\tau\_n \nearrow \infty$ a sequence of positive numbers. 


I can't demonstrate that the following sequence of measures is or is not convergent in the weak topology:
$$\frac{1}{\tau\_n} \int\_0^{\tau\_n} \mu(A-t) dt.$$


Can someone help me?


Thank you!



Is the following sequence convergent in the weak topology?
Let $\mu\_n$ be the measure defined by $$\mu\_n(A) = \frac{1}{\tau\_n} \int\_0^{\tau\_n} \mu(A-t)\,dt.$$


In general, this sequence doesn't converge weakly. Suppose for instance that $\mu$ is a point mass at $0$. Then $\mu\_n$ is the uniform probability measure on $[0, \tau\_n]$. Let $f$ be any bounded, continuous, strictly positive, Lebesgue integrable function on $\mathbb{R}$ (for instance, $f(x) = e^{-x^2}$ would do). Then 
$$\int\_\mathbb{R} f\,d\mu\_n = \frac{1}{\tau\_n} \int\_0^{\tau\_n} f(t)\,dt \le \frac{1}{\tau\_n} \int\_{\mathbb{R}} |f(t)|\,dt \to 0$$
as $\tau\_n \to \infty$. Since each $\mu\_n$ is a positive measure, if there is a weak limit $\mu^\*$ it is also a positive measure. So this would force $\int f\,d\mu^\* = 0$, which can only happen if $\mu^\* = 0$.


But $\mu\_n$ doesn't converge to 0 either; take $f=1$. So the sequence doesn't converge weakly at all.


If you had $\tau\_n \to 0$ (which is how I originally misread the question) then $\mu\_n \to \mu$ weakly:


1. Verify that $\mu\_n$ is indeed a measure (i.e. that it is countably additive). Use the monotone convergence theorem.
2. 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").
3. Use Fubini's theorem to verify that
$$\tag{\*} \int\_{\mathbb{R}} f\,d\mu\_n = \int\_{\mathbb{R}} \frac{1}{\tau\_n} \int\_0^{\tau\_n}f(x+t)\,dt\,\mu(dx).$$
4. 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 $\|f\|\_\infty$.
5. 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.


