Let $(X,\mathcal{F},\mu)$ be a measure space. Let $f\in\mathscr{L}^{1}(X,\mathcal{F},\mu)$. Then $\forall\epsilon>0 \ \exists\delta>0\text{ s.t. }\forall A\in\mathcal{F}$ $$\mu(A)\le\delta \implies \int\limits _{A}|f| \, d\mu\le\epsilon$$i.e. The Lebesgue Integral is absolutely continuous w.r.t. the Lebesgue Measure $$\int\limits \, d\mu\ll\mu$$