Here my book has given a proof. This is part of a larger proof on Lebesgue integrability criterion. I am trying to determine how from $w(f;c) = 0$ we get that $\forall \epsilon \gt 0$ $\exists s \gt 0$ with $W(f; V_c(c)) \lt \epsilon$.
I know that since it is assumed that, $w(f;c) =0$, by definition: $inf \{W(f;V_s(c):s>0)\}=Lim_{s \to 0^+} (W(f;V_s(c))) = 0$.
So from this I have that $Lim_{s \to 0^+} (W(f;V_s(c))) =Lim_{s \to 0^+} (W(f;x \in A: |x-c|<s) )) = 0$.
So from here I used the limit definition since the limit is $0$. So: $\forall\epsilon \gt 0 \exists \delta \gt0$ such that if $0<|x-c|< \delta$ then $|f(x)|<\epsilon$. Fixing $\epsilon > 0$, letting $\delta = s$, I have that if $0<|x-c|<s$ then $|f(x)| < \epsilon$.
I'm stuck here and need some assitance figuring out how the second statement can be true. Any help is appreciated.
Since $\lim_{t\to 0^+}W(f; V_t(c))=0$, by definition of the limit this means that for all $\epsilon>0$, there exists a $t>0$ such that for all $0< s<t$ we have $|W(f; V_s(c))|<\epsilon$. In particular, there is at least one such $s$ (e.g. $s=t/2$) to which this applies, which is what you wanted.