$A\subset \mathbb R$ is said to be Lebesgue measurable if for each $E\subset \mathbb R$,$m^*(E)=m^*(E\cap A)+m^*(E\cap A^c)$ where $m^*$ is the Lebesgue outer measure.Now I have to prove that the following are equivalent:
$1. A$ is Lebesgue measurable.
$2.\exists $a Borel set $B\subset A$ such that $m^*(A-B)=0$.
$3. $For each $\epsilon>0$ there exists a closed set $F_\epsilon\subset A$ such that $m^*(A-F_\epsilon)<\epsilon$.
$4.\exists $ open sets $G_1,G_2,...\supset A$ such that $m^*(\bigcap\limits_{n=1}^\infty G_n-A)=0$.
$5.$There exist closed ets $F_1,F_2,...\subset A$ such that $m^*(A-\bigcup\limits_{n=1}^\infty F_n)=0$.
$6.$ There exists a Borel set $B\supset A$ such that $m^*(B-A)=0$.
$7.$ For each $\epsilon>0$ there exists an open set $G_\epsilon\supset A$ such that $m^*(G_\epsilon-A)<\epsilon$.
I could prove the following implications $(1)\implies (7)\implies (6)$ and $(3)\implies (5)\implies (2)$ and $(7)\implies (4)\implies (6)$ but I am having problem with the others.
Can someone show me a proper route so that I can prove that all of these are equivalent?Also,as I am a beginner,can someone show me how to prove those implications?
Addendum
I had asked one question earlier but that did not show me all these are equivalent.
You say you have shown $(1) \implies (7) \implies (4) \implies (6)$. To finish, it will suffice to show $(6) \implies (1)$ and then relate the remaining statements to these by taking complements.
Here is a proof of $(6) \implies (1)$:
Suppose $B$ is a Borel set such that $B \supset A$ and $m^*(B \setminus A) = 0$, and let $E \subset \mathbb{R}$ denote any set. Since Borel sets are Lebesgue measurable, $$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c) \\ m^*(E\cap A^c) = m^*(E\cap A^c\cap B) + m^*(E\cap A^c\cap B^c)$$ Considering the second of these equations, we note $$E\cap A^c \cap B^c = E\cap B^c \\ E\cap A^c \cap B \subset B\setminus A \implies m^*(E\cap A^c \cap B) = 0,$$ so that $$m^*(E\cap A^c) = m^*(E\cap A^c\cap B) + m^*(E\cap A^c\cap B^c) = m^*(E\cap B^c).$$ Additionally, since $$E\cap A \subset E\cap B \subset (E\cap A)\cup (B \setminus A) \implies m^*(E\cap A) \leq m^*(E\cap B) \leq m^*(E\cap A) + m^*(B\setminus A),$$ we have that $m^*(E\cap A) = m^*(E\cap B).$ Putting it together, $$m^*(E) = m^*(E\cap B) + m^*(E\cap B^c) = m^*(E\cap A) + m^*(E\cap A^c),$$ and since $E\subset \mathbb{R}$ was arbitrary, $A$ is Lebesgue measurable.
Combined with what you'd already shown, this shows $(1), (4), (6), (7)$ are equivalent statements. Now, instead of writing separate proofs for the equivalence of the others, we use the fact that $(1)$ is equivalent [which is nearly obvious by your definition] to $$(1^c) \quad A^c \text{ is Lebesgue measurable}$$
By the equivalence of $(1), (4), (6), (7)$, it follows that $(1^c)$ is equivalent to each of $(4^c), (6^c), (7^c)$ where $(n^c)$ is the result of replacing $A$ by $A^c$ in $(n)$. By DeMorgan's laws, the duality between open/closed sets, and the fact the Borel sets are closed under complements, we can identify $(4^c)$ with $(5)$, $(6^c)$ with $(2)$, and $(7^c)$ with $(3)$.
In doing this, we conclude $(1)-(7)$ are equivalent statements.