Im trying to prove the next:
Let $(X,\mathcal{S},\mu)$ a measure space and let $\mathcal{F}=\{A\in\mathcal{S}:\mu(A)<\infty\}.$ Defining a relation ~ in $\mathcal{F}\times\mathcal{F}$ as: $A$~$B$ iff $\mu(A\triangle B)=0.$ Prove that:
i) ~ is an equivalence relation,
ii) If $\overline{\mathcal{F}}$ is the resulting quotient space by such relation, then $d:\overline{\mathcal{F}}\times\overline{\mathcal{F}}\rightarrow\mathbb{R}$ given by $d([A],[B])=\mu(A\triangle B)$ is a well defined metric and $(\overline{\mathcal{F}},d)$ is a complete metric space.
iii) If $(X,\mathcal{S},\mu)$ is the Lebesgue measure space over real numbers, then $(\overline{\mathcal{F}},d)$ is path-connected. (Hint: Let $\gamma:[0,\infty)\rightarrow\overline{\mathcal{F}}$ defined by $\gamma(s)=[A\cap(-s,s)].$ Then $\gamma$ is continuous, $\gamma(0)=[\emptyset],$ and $d(\gamma(s),[A])\rightarrow 0$ when $s\rightarrow\infty.$)
iv) $(\overline{\mathcal{F}},d)$ could not be locally compact even if $\mu(X)<\infty.$
I've already proved i) and ii). To prove iii) I'm following the suggestion given; I've proved such things but I cannot see why this implies or helps to get the desired result.
For iv) I was thinking about $X=[0,1],$ $S=\mathbb{B}_{[0,1]}$ and $\mu$ the Lebesgue measure and considering $A_n=\displaystyle\bigcup_{k=1}^{2^{n-1}}\left[\frac{2k-1}{2^n}\frac{2k}{2^n}\right].$ So $d([A_n],[A_m]=\frac{1}{2}$ when $n\neq m.$ So we can conclude that $(\overline{\mathcal{F}},d)$ is not compact, but I don't know how to prove is not locally compact.
Any kind of help is thanked in advanced.
Define $\gamma_1$ on $[0,\pi/2]$ by $\gamma_1 (t) =\gamma (\tan\, t)$ if $t <\pi/2$ and $\gamma_1(\pi /2)=[A]$. This gives path from $[\emptyset ]$ to $[A]$ for any $A$. Hence the space is path connnected.
Part iv): Let $\epsilon >0$ and $B_n=\epsilon A_n (\equiv \{\epsilon t: t\in A_n\})$ where $A_n$ are the sets you have constructed. Then the sequence $([B_n])$ is contained in the $\epsilon$ ball around $[\emptyset ]$ and it has no convergent subsequence. It follows that no neighborhood of $[\emptyset ]$ is relatively compact. Hence the space is not locally compact.