Proving normal distribution is a probability measure/ Sigma additivity

Question

Let $$P_X(B)=\mathbb{P}(X\in B)=\int_B \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma)^2}\,\lambda(dx)$$

$B$ borel and with $\mu$ and $\sigma$ parameters. Prove that $P_X$ is a probability measure.

The 3 steps are straight forward but I don't know how to prove sigma additivity

1. $P_X$ is defined on a Sigma algebra.
2. $$\int_\emptyset \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma)^2}\,\lambda(dx) = \frac{1}{\sqrt{2\pi}\sigma}\int_\emptyset e^{-(x-\mu)^2/(2\sigma)^2}\,\lambda(dx) \leq \lambda(\emptyset)=0$$

3. $P_X(\mathbb{R})=1.$ I know how to show this with the variable transformation.

4. How to show Sigma additivity?

Answer 1

Let $B_1,B_2,\dots$ be disjoint Borel measurable subsets of $\mathbb R$.

Let us abbreviate the integrand with $f$ which is a nonnegative and measurable function (that satisfies $\int f(x)\lambda(dx)=1$).

By definition:

$$P_X(\bigcup_{n=1}^{\infty}B_n)=\int1_{\bigcup_{n=1}^{\infty}B_n}(x)f(x)\lambda(dx)$$

Now observe that $1_{\bigcup_{n=1}^{\infty}B_n}(x)f(x)=\lim_{k\to\infty} f_k(x)$ where $f_k(x)=\sum_{n=1}^k1_{B_n}(x)f(x)$.

This with $0\leq f_1\leq f_2\leq\cdots$ so applying MCS we find $$P_X(\bigcup_{n=1}^{\infty}B_n)=\lim_{k\to\infty}\int f_k(x)\lambda(dx)=\lim_{k\to\infty}\sum_{n=1}^k\int1_{B_n}(x)f(x)\lambda(dx)$$$$=\lim_{k\to\infty}\sum_{n=1}^kP_X(B_n)=\sum_{n=1}^{\infty}P_X(B_n)$$

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edit: smart alternative using that $\mathbb P$ denotes a (probability) measure.

Let $B_1,B_2,\dots$ be disjoint Borel measurable subsets of $\mathbb R$.

Then the sets $\{X\in B_1\},\{X\in B_2\},\cdots$ are disjoint measurable sets and since $\mathbb P$ denotes a (probability) measure we conclude:$$P_X(\bigcup_{n=1}^{\infty}B_n)=\mathbb P(X\in\bigcup_{n=1}^{\infty}B_n)=\mathbb P(\bigcup_{n=1}^{\infty}\{X\in B_n\})=\sum_{n=1}^{\infty}\mathbb P(X\in B_n)=\sum_{n=1}^{\infty}P_X(B_n)$$