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Let $\mu_n = \mathcal{N}(0,n)$ be the normal distribution with mean $0$ and variance $n$ on $\mathbb{R}$, $\nu$ the zero-measure (which is defined by $\nu(A) = 0$ for any $A\in\mathcal{B}(\mathbb{R})$), $\lambda$ the Lebesgue measure on $\mathbb{R}$.

Show that $\mu_n \overset{\mathrm{v}}{\rightarrow} \nu$, but the sequence $(\mu_n)$ does not converge weakly.

Also show that $\sqrt{2 \pi n}\mu_n \overset{\mathrm{v}}{\rightarrow} \lambda$.

Note: "$\overset{\mathrm{v}}{\rightarrow}$" denotes vague convergence, "$\overset{\mathrm{w}}{\rightarrow}$" denotes weak convergence.

To show that $\mu_n \overset{\mathrm{v}}{\rightarrow} \nu$, we have to show that $$ \int f \, d\mu_n \overset{n\rightarrow\infty}{\longrightarrow} \int f \, d\nu = 0 \quad \text{for any $f \in C_c(\mathbb{R})$}$$

If $f \in C_c(\mathbb{R})$, $f$ is also bounded, so if $A$ is the support of $f$ we have $$ |f(x)| \leq \|f\|_\infty 1_A (x) \quad \text{for all } x\in\mathbb{R}$$ Then \begin{align*} \Bigl|\int f \, d\mu_n \Bigr| &\leq \int |f|\, d\mu_n \leq \int \|f\|_\infty 1_A \, d\mu_n = \frac{1}{\sqrt{2 \pi n}}\int_A \|f\|_\infty \exp\Bigl(-\frac{x^2}{2n}\Bigr) \, dx \\ & \leq \frac{1}{\sqrt{2\pi n}}\int_A \|f\|_\infty \, dx \overset{n\rightarrow\infty}{\longrightarrow} 0 \, , \end{align*} which proves that $\mu_n \overset{\mathrm{v}}{\rightarrow} \nu$.

If we take the function $1 \in C_b(\mathbb{R})$ we get\begin{align*} \int 1 \, d\mu_n = \frac{1}{\sqrt{2\pi n}}\int_{-\infty}^\infty 1 \cdot \exp\Bigl(-\frac{x^2}{2 n}\Bigr) \, dx = 1 \, . \end{align*} Assume there were a measure $\mu$ so that $\mu_n \overset{\mathrm{w}}{\rightarrow} \mu$. $\mu$ has to be consistent with what we already found: \begin{equation*} \mu(A) = 0 \text{ if $A$ compact} \quad \mu(\mathbb{R}) = 1 \, . \end{equation*} Let $A_i = [i, i+2]$. The $A_i$ are compact and $\bigcup_{i=1}^\infty A_i = \mathbb{R}$. Since $\mu$ is a measure, it has to be $\sigma$-subadditive, but now we get a contradiction: $$ 1 = \mu(\mathbb{R}) = \mu\Bigl(\bigcup_{i=1}^\infty A_i\Bigr) \leq \sum_{i=1}^\infty \mu(A_i) = 0 \, .$$

Now we want to show that $\sqrt{2 \pi n}\mu_n \overset{\mathrm{v}}{\rightarrow} \lambda$, i. e.: \begin{align*} \int f \sqrt{2 \pi n}\, d\mu_n &\overset{n\rightarrow\infty}{\longrightarrow} \int f \, d\lambda \quad \text{for any $f \in C_c(\mathbb{R})$}\\ \end{align*} which is equivalent to \begin{align*} \int f(x) \exp\Bigl(-\frac{x^2}{2n}\Bigr) \, dx &\overset{n\rightarrow\infty}{\longrightarrow} \int f(x) \, dx \quad \text{for any $f \in C_c(\mathbb{R})$}\, . \end{align*} Since $|1-\exp(x^2/2n)|\leq 1$ and $f \in C_c(\mathbb{R})$ there exists an integrable majorant $$\Bigl|f(x)\Bigl(1-\exp\Bigl(-\frac{x^2}{2n}\Bigr)\Bigr)\Bigr| \leq \|f\|_\infty 1_A \quad \text{(as above $A$ is the support of $f$)}$$ so we can use the dominated convergence theorem: $$ \lim_{n\rightarrow\infty} \biggl|\int f(x)\Bigl(1-\exp\Bigl(-\frac{x^2}{2n}\Bigr)\Bigr) \, dx\biggr| \leq \int\lim_{n\rightarrow\infty}\biggl| f(x)\Bigl(1-\exp\Bigl(-\frac{x^2}{2n}\Bigr)\Bigr)\biggr| \, dx = \int 0 \, dx = 0\, ,$$ which proves that $\sqrt{2 \pi n}\mu_n \overset{\mathrm{v}}{\rightarrow} \lambda$.

Is everything correct? I'll be eternally grateful to the person who checks my proof!

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Since $|1-\exp(x^2/2n)| \leq 1$

You mean $\exp(\color{red}{-}x^2/2n)$, I suppose.

so we can use the dominated convergence theorem

Your application of the dominated convergence theorem is correct. However, from my point of view, the following argumentation is more direct/easier to follow (but that's primarily oppinion-based):

Since $|\exp(-x^2/2n)| \leq 1$ and $f \in C_c$, there exists an integrable majorant $$\left| f(x) \exp \left(- \frac{x^2}{2n} \right) \right| \leq \|f\|_{\infty} 1_A(x).$$ It follows from the dominated convergence theorem that $$\lim_{n \to \infty} \int f(x) \exp \left(- \frac{x^2}{2n} \right) \, dx = \int f(x) \, dx.$$

Everything else is correct and well-written.

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  • $\begingroup$ Yes exactly, I meant $\exp(-x^2/2n)$. Regarding the step where I apply the dominated convergence theorem, I also put the absolute value $ | \cdot |$ inside the integral, but didn't write it explicitly in a separate step, therefore I thought it has to be an inequality. $\endgroup$ – Qyburn Apr 19 '15 at 19:58
  • $\begingroup$ @Qyburn You are right. See my edited answer. $\endgroup$ – saz Apr 19 '15 at 20:07
  • $\begingroup$ I agree that it is clearer that way. Thank you again! $\endgroup$ – Qyburn Apr 22 '15 at 3:34

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