I am stuck on understanding one point Proposition 11 (Real Analysis by Royden and Fitzpatrick, 4th edition, Page-66)
Proposition 11: Let$ f$ be a simple function defined on a set E of finite measure. Then for each $ε>0$, there is a continuous function g on R and a closed set F contained in E for which $f=g$ on F and $m(E−F)<ε$.
Proof(in short): Let $a_1,a_2,…,a_n$ be the finite number of distinct values taken by $f$, and let them be taken on the sets $E_1,E_2,…,E_n$ respectively. The collection ${{E_n}}_{n=1}^{k}$ is disjoint since the $a_k$'s are distinct. According to Theorem 11 of Chapter 2, we may choose closed sets $F_1,F_2,…,F_n$ such that for each index $k,1≤k≤n$, $F_k⊆E_k$ and $m(E_k−F_k)<ε/n$. Define $ g$ on $F$ to take the value $a_k$ on $F_k$ for $1≤k≤n$. Since the collection ${F_k}_{k=1}^{n} $ is disjoint, $g$ is properly defined
Now $m(E-F) =m(\bigcup_{i=1}^{n} [E_k-F_k])=\sum_{n=1}^{n} {(E_k−F_k)<ε}$
The Part i am having trouble is Assuming $ m(E_k−F_k)<ε/n$ why $\sum_{n=1}^{n} {(E_k−F_k)<ε} $ ?
How $ε+ ε/2+ ε/3 +.....=ε/n<ε?$
You are mixing the index with the number of elements. The sum is $$ \sum_{k=1}^n\frac{\varepsilon}n=\frac{\varepsilon}n\,\sum_{k=1}^n1=\varepsilon. $$