Definition of measurable set: A set $E$ measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.
Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.
Is proving $m(E) < \delta, \forall \delta > 0$ equivalent to prove $m(E) = 0$? $m$ denotes Lebesgue measure on measurable set.)
I have this question when I was reading Lusin theorem, that is, if $f(x)$ is a Lebesgue measurable function and finite almost everythere($m({x \in E: |f(x)| = +\infty}) = 0$), then for $\forall \delta > 0, \exists F \subset E$ where $F$ is closed, such that $f(x)$ is continuous on $F$ and $m(E - F) < \delta$. Lusin theorem can fail to work if $m(E - F) < \delta$ replaced by $m(E - F) = 0$. This made me confused coz sometimes I used $m(E) < \epsilon, \forall \epsilon > 0$ to show $m(E) = 0$ in my homework which was introduced by my professor.
What's wrong with it?
It is true that $$ m(E) = 0 \quad \iff \quad \forall \epsilon > 0 \, : \, m(E) < \epsilon. $$ This has nothing to do with measure theory; it's a property of real numbers that if $x \ge 0$ is less than any positive number, then $x = 0$.
However, the statement of Lusin's theorem is not this. Lusin's theorem says $$ \forall \delta > 0 \; \exists F \subset E \; : \; F \text{ is closed, } f \text{ is continuous on } F\text{, } m(E \setminus F) < \delta. \tag{1} $$ Note that $F$ depends on $\delta$; this is is NOT the same as $$ \exists F \subset E \; \forall \delta > 0\; : \; F \text{ is closed, } f \text{ is continuous on } F\text{, } m(E \setminus F) < \delta. \tag{2} $$ In (2), you can conclude that $m(E \setminus F) = 0$, but in (1) you cannot.