Weak absolute continuity of measures

Question

I want to show that if we have a function $f \in L^p$ sucht that $||f||_p =1$.

Define a new measure $\mu$ by

$$\mu(A):=\int_A |f(x)|^p dm(x).$$

Then $\forall \epsilon > 0 \ \ \exists \delta>0$ such that the following implication holds:

$$m(A)< \delta \Rightarrow \mu(A) < \epsilon.$$

I read a few things in wikipedia and noticed that this is somehow related to the absolute continuity of measures, but I don't know anything about this. Maybe, you can give me a hint how to show this?

Answer 1

1. Considering $|f|^p$ instead of $f$, we can assume that $p=1$ and $f$ is non-negative.

2. For each simple function $s:=\sum_{i=1}^Na_i\chi_{B_i}$, where $a_i$ are non-negative real numbers and $B_i$ disjoint measurable sets, and each measurable set $A$, $$\mu(A)=\int_A|f-s+s|\mathrm m(x)\leqslant \int_A|f-s|\mathrm dm(x)+\int_As(x)\mathrm dm(x)\\\leqslant\lVert f-s\rVert_1+\left(\sum_{i=1}^Na_i\right)m(A)\tag{*}.$$

3. We can conclude by $(*)$ by a $2\varepsilon$-argument: take $s$ such that $\lVert f-s\rVert_1\lt \varepsilon$; then define $\delta:=\frac{\varepsilon}{1+\sum_{i=1}^Na_i}$.