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I'm trying to prove that a countable group $G$ being amenable, implies that it satisfies the Folner condition.

There are some steps I need to prove. Among them:

1: For every $r \in [0,1]$, we denote by $E_r$ the function that is equal to $1$ on $(r, 1]$ and equal to $0$ elsewhere. Prove that for every $a, b \in [0,1]$ we have $$ | a - b | = \int_0^1 | E_r(a) - E_r(b)| dr \quad \text{and} \quad a = \int_0^1 E_r(a) dr. $$

2: Whenever $\xi: G \rightarrow [0,1]$ and $r \in [0,1]$, we denote by $\xi^r := E_r \circ \xi$. Note that $\xi^r = \chi_{A_r}$ where $\chi$ is the indicator function and $A_r = \left\{ g \in G \mid \chi(g) \in (r, 1] \right\}$. Prove that for all finitely supported functions $\xi, \eta: G \rightarrow [0,1]$, we have $$ || \xi - \eta ||_1 = \int_0^1 || \xi^r - \eta^r ||_1 dr \quad \text{and} \quad || \xi||_1 = \int_0^1 || \xi^r ||_1 dr. $$

Attempt: I'm not sure how to solve this. For $a \in [0,1]$,I wish to show that $a = \int_0^1 E_r(a) dr$. Since $E_r$ equals $1$ on $(r,1]$ and zero elsewhere, I think that $\int_0^1 E_r(a) dr = \int_r^1 E_r(a) dr. $ So now we know that $a \in (r,1]$. Now I wanted to make different case, but I don't get the desired result.

Also not sure how to solve 2).

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