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

Question

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 1

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

Answer 2

This really should be a comment, but it's a bit too long.

Part 2 is very similar to part 1. $$\begin{aligned} \int_A|f_k| &= \int_{A \cap [|f_k| \geq h]}|f_k| + \int_{A \setminus [|f_k| \geq h]} |f_k| \\ &\leq \int_{A \cap [|f_k| \geq h]}|f_k| + \int_{A \setminus [|f_k| \geq h]} h \\ &\leq \int_{[|f_k| \geq h]}|f_k| + \int_{A} h \end{aligned}$$ where the last inequality follows because the integrands are nonnegative and both sets increase from line 2 to line 3.