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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!

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  • $\begingroup$ Hint: If $\mu_n$ is the measure you defined, can you show that $$\int f\,d\mu_n = \frac{1}{\tau_n} \int_0^{\tau_n} \int_{\mathbb{R}} f(x + t)\,\mu(dx)\,dt$$ for bounded continuous $f$? Then try Fubini and dominated convergence. $\endgroup$ – Nate Eldredge Feb 11 '17 at 17:42
  • $\begingroup$ I don't understand. Can you give me more details, please? Thank you! $\endgroup$ – g.pomegranate Feb 11 '17 at 17:57
  • $\begingroup$ "Weak topology" in this context means different things to probabilists and analysts, so can you please state precisely which topology is meant? $\endgroup$ – Nate Eldredge Feb 11 '17 at 17:59
  • $\begingroup$ I want to verify if there is a measure $\mu^*$ such that $$(f, \mu_n) \to (f, \mu^*),$$ for $f$ bounded and continuous, where $(f, \mu)$ means $\int f d\mu$. $\endgroup$ – g.pomegranate Feb 11 '17 at 18:03
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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.

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  • $\begingroup$ Hold on a minute! Surely the convergence is just to 0. If $f$ is any compactly supported function, then $|\int f(x+t)\,d\mu|<\epsilon$ for all $t$ outside a bounded range. $\endgroup$ – Anthony Quas Feb 12 '17 at 9:08
  • $\begingroup$ @AnthonyQuas: Oh, I didn't read the question carefully and I was thinking of $\tau_n \to 0$. But note that $f$ is not assumed to be compactly supported. Let me look at this again when I have more time. $\endgroup$ – Nate Eldredge Feb 12 '17 at 15:47
  • $\begingroup$ I think this is convolution of a uniform probability measure on $[0,\tau_n]$ with $\mu$; this sequence isn't tight and converges to the 0 measure. $\endgroup$ – Anthony Quas Feb 12 '17 at 20:54
  • $\begingroup$ @AnthonyQuas: In this topology, it doesn't converge to the zero measure; see above. (You might be thinking of the weak-* topology on $C_0(\mathbb{R})$.) $\endgroup$ – Nate Eldredge Feb 12 '17 at 21:45
  • $\begingroup$ OK. Your answer seems exactly right. $\endgroup$ – Anthony Quas Feb 15 '17 at 9:19

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