3

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!

JJJ
  • 874
  • What is $g$? What does $\in>$ mean? – David C. Ullrich Sep 01 '15 at 22:31
  • sorry typos, should be $h$ – JJJ Sep 01 '15 at 22:32
  • 1
    Apply part 2, except with $\epsilon/2$ in place of epsilon. Now you have $\epsilon/2 + \delta$. If only $\delta$ were smaller than $\epsilon/2$. You can make $\delta<\epsilon/2$ just by adding that to whatever other conditions you imposed on $\delta$... – David C. Ullrich Sep 01 '15 at 22:37
  • hmm, what do you mean 'whatever other conditions you imposed on $\delta$'. Do you mean just say that I choose a $\delta < \epsilon /2 $? Thanks for your help! – JJJ Sep 01 '15 at 22:42
  • Yes. If for every $\epsilon>0$ there exists $\delta$ such that $A<\delta$ implies $B<\epsilon$ then it follows that for every $\epsilon>0$ there exists $\delta>0$ such that $\delta<\epsilon/2$ and such that $A<\delta$ implies $B<\epsilon/2$. – David C. Ullrich Sep 01 '15 at 23:38
  • Sorry David, I don't quite understand what you mean. Now given $ \epsilon $ I know that there is an $h$ such that $ \int_{[|f_k| \ge h]} |f_k| < \epsilon $, but then how can I construct a bounds on $h$? I'm not sure how I can draw any conclusions on $h$? – JJJ Sep 02 '15 at 02:23
  • You said you could do part 3 except you got $\epsilon+\delta$ instead of $\epsilon$. I was explaining how if you got $\epsilon+\delta$ you automatically got $\epsilon$. – David C. Ullrich Sep 02 '15 at 14:03
  • Sorry I did not write that very clearly. I don't really know how to do part 3, except that on the LHS of part 2, one integral I can choose to be bounded by $\epsilon$, and the other if $ < \delta $ provides a bound on $\int_A |f_k|d\mu $ - but I'm not sure how I can ensure $\int_A h d\mu < \delta$ – JJJ Sep 02 '15 at 23:12

2 Answers2

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

Eugene Zhang
  • 16,805
  • Thanks for your answer. So have you proved in another post that for any $ \epsilon $ there is $ \eta$ such that $\mu(A) < \eta \implies \int_A h d\mu < \epsilon $? I am trying to read that - but it's a bit hard to follow. – JJJ Sep 02 '15 at 08:27
  • Yes, I proved it. It should not be too hard. This property is known as absolute continuity of Lebesgue integral. – Eugene Zhang Sep 02 '15 at 08:33
  • I think I am starting to get it, so can I just confirm - $h$ is determined by our choice of $\epsilon$. But once we have our $h$, we can choose a second $ \epsilon $, which determines $\eta$ and whenever $\mu(A) < \eta $ then we have our result. So $A$ depends on $\eta$ which depends on $\epsilon$ – JJJ Sep 02 '15 at 10:02
  • Yes you got it Right – Eugene Zhang Sep 02 '15 at 17:18
0

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.