Proving inequality Lebesgue integral

Question

Let $f \in L^{1}(\mu). \ \forall n\geq 1,$ let $f_n=\min(|f|,n).$ Let $\varepsilon>0.$

1. Show that there exist an $n$ such that $$0\leq \int (|f|-|f_n|)d\mu <\frac{\varepsilon}{2}.$$

2. Prove the existence of a $\delta >0 $ such that if $\mu (A)<\delta$, then

$$\int_A|f| d\mu<\varepsilon.$$

Ideas.

I only have ideas about the proof of 1., but I'm not sure if they are useful and and I don't know how to conclude, I would appreciate suggestions about both proofs.

For the inequality 1. \begin{eqnarray} \int f_nd\mu &\leq& \int|f|d\mu\\ 0&\leq& \int|f|d\mu-\int f_nd\mu\\ 0&\leq& \int(|f|-|f_n|)d\mu. \end{eqnarray}

And I wonder if $f_n$ converges almost everywhere to $|f|$?, and i think that this may be useful.

Answer 1

As you have already mentioned, you know that $0 \leq f_n \leq |f|$. Moreover, since $f \in L^1(\mu)$, we know that $|f| < \infty$ $\mu$-almost everywhere. Thus $f_n \rightarrow f$ $\mu$-almost everywhere and the convergence is monotone. By the montone convergence theorem (or Lebesuge dominated convergence theorem) you can now conclude (1). Because of $||f|-f_n| = |f|-f_n$ you also have $L^1$ convergence.

(2) can be proven with the help of (1): Note that $$\int_A |f| d \mu \leq \int (|f| - f_n )d \mu + \int_A f_n \mu < \varepsilon/2 + \int_A f_n d \mu. $$ Thus, the statement will be proven if we show (2) for bounded functions. This case is simple: Take $\delta = \varepsilon/\|f\|_\infty$, then one has for every $\mu(A) < \delta$ that $$\int_A |f| d \mu \le \|f\|_\infty \mu(A) < \|f\|_\infty \delta = \varepsilon.$$

Answer 2

The first part of the question can be easily proved by monotone convergence.

By the first part exists $m$ such that $\int|f|-f_m<\epsilon$

Chose $\delta=\frac{\epsilon}{2m}$ and $A$ such that $\mu(A)<\delta$

Then $$\int_A|f|=\int_Af_m+\int_A|f|-f_m \leq \int_Af_m+\int |f|-f_m$$ $$\leq m \mu(A)+\epsilon/2 <\epsilon$$