measure theory - Show that for every $epsilon > 0$ there exists $h in mathcal{L}^1(X)$ non-negative and $delta > 0$ such that:

I am working through some practice questions, and I think I have gotten the first two parts, but I am having trouble deriving the third part:

Let $(X,\mathcal{A},\mu)$ be a finite measure space. Suppose that
$(f_k)$ is a sequence of measurable functions $X \rightarrow \mathbb{R}$ such that for every $\epsilon > 0$ there exists $h \in \mathcal{L}^1(X)$ non-negative such that:

$$\int_{[|f_k|\ge h]} |f_k|d\mu< \epsilon$$

for all $k \in \mathbb{N}$. Where $[|f_k|\ge h] = \{ x \in X : |f_k(x)| \ge h(x) \}$

(1) Show that there exists $P>0$ such that:

$$\int_X |f_k|d\mu \le P$$ for all $k \in N$

(2) Show that for every $A \in \mathcal{A}$ and every $h \in > \mathcal{L}^1(X)$ non-negative:

$$\int_A |f_n|d\mu \le \int_{[|f_k|\ge h]} |f_k|d\mu + \int_A h d\mu$$
(3) Using part (2), show that for every $\epsilon > 0$ there exists $h \in \mathcal{L}^1(X)$ non-negative and $\delta > 0$ such that:

$A \in \mathcal{A}$ and $\int_A h d\mu < \delta \implies \int_A |f_k|d \mu < \epsilon$ for all $n \in \mathbb{N}$

For part (1), I have written the integral on the left hand side as disjoint integrals, namely $[|f_k|\ge h]$ and $[|f_k| < h]$ then the second integral is smaller than $\int_{[|f_k| < h]} h$, since it is precisely over the x's which $h > |f_k|$. And since we know the integrals of $h$ are finite, this yields the result.

For part (2), I have done a similar construction, splitting the problem into two cases, where $A$ and $[|f_k|\ge h]$ intersect and where they do not. I am able to derive the inequalities. Is this the right approach to this problem?

Part(3) is where I am having the most trouble, by part(2) it seems that I can immediately derive that $\int_A |f_k|d\mu < \epsilon + \delta$, but how to show it is just $< \epsilon$?

Any help would be very gratefully received!

Answer

Part 3: Given $\epsilon>0$, there is $h$ such that

$$\int_{[|f_k|\ge h]} |f_k|d\mu< \epsilon$$
For this $h$, by post for each $\epsilon >0$ there is a $\delta >0$ such that whenever $m(A)<\delta$, $\int_A f(x)dx <\epsilon$,

given $\epsilon$, there is a $\eta$, for any $A$ such that $\mu(A)<\eta$, there is

$$\int_A h d\mu < \epsilon$$

So
$$\int_A |f_n|d\mu=\int_{A\cap [|f_n|\ge h]} |f_n|d\mu+\int_{A\cap [|f_n|< h]} |f_n|d\mu\leqslant \int_{[|f_n|\ge h]} |f_n|d\mu+\int_A h d\mu<2\epsilon$$