Draft

Let $(X,d)$$(X,d)$ be a pseudo-metric space, we could do the Metric Quotient. The topology basis is $B_a(r)=\{x\in X:d(x,a)<r\}$$B_a(r)=\{x\in X:d(x,a)< r\}$.

We define $x\sim y⟺d(x,y)=0$$x\sim y\iff d(x,y)=0$, we get a quotient map $\pi :X\to X/\sim$$\pi:X\to X/\sim$. Then we consider the metric $d_X([x],[y])=d(x,y)$$d_X([x],[y])=d(x,y)$, whcih is a metric. Easy to see that $\pi :X\to X/\sim$$\pi:X\to X/\sim$ is continuous.

Let $f:(X,d)\to (Y,D)$$f:(X,d)\to (Y,D)$ be a continupus function, where $(Y,D)$$(Y,D)$ is a metric space. By the universal property of quotient topology we have:

Corollary. $f:(X,d)\to (Y,D)$$f:(X,d)\to (Y,D)$ is continuous at $x_0$$x_0$ iff

$$\forall \epsilon >0,\exists \delta >0,d(x,x_0)<\delta ⟹D(f(x),f(x_0))<\epsilon .$$

Now let $(M,\mathrm{\Sigma },\mu )$$(M,\Sigma,\mu)$ be a measure space with $\mu (M)<\mathrm{\infty }$$\mu(M)<\infty$.

We could define a pseudo metric on $\mathrm{\Sigma }$$\Sigma$, given by $d_{\mathrm{\Sigma }}(A,B)=\mu (A\mathrm{\Delta }B)$$d_{\Sigma}(A,B)=\mu(A\Delta B)$.

$$\mu (A\mathrm{\Delta }C)=\mu (A\mathrm{\Delta }B\mathrm{\Delta }B\mathrm{\Delta }C)\le \mu (A\mathrm{\Delta }B)+\mu (A\mathrm{\Delta }C)$$

Proposition. $\mu :\mathrm{\Sigma }\to {\mathbb{R}}^{+}$$\mu:\Sigma\to\mathbb R^+$ is continuous.

Proof. $|\mu (A)-\mu (B)|=|\mu (A-B)-\mu (B-A)|\le \mu (A-B)+\mu (B-A)=\mu (A\mathrm{\Delta }B)◻$$|\mu(A)-\mu(B)|=|\mu(A-B)-\mu(B-A)|\le\mu(A-B)+\mu(B-A)=\mu(A\Delta B)\quad\square$

Corollary. Let $(A_n)$$(A_n)$ be a convergent sequence such that $A_n\to A$$A_n\to A$, then we have $\underset{n\to \mathrm{\infty }}{lim}\mu (A_n)=\mu (A)$$\lim_{n\to\infty}\mu(A_n)=\mu(A)$.