MATH 597

MATH 597: Measure Theory โ€” Supplementary Material

Homework 0, problem-solving lectures, and practice problems

Qiulin Fan ยท Winter 2025

hw 0

(Not graded.) Read Sections 0.1-0.3 and 0.5-0.6 in Follandโ€™s book. Note: I expect you to have seen much but not necessarily all of this material in earlier courses. It is not necessary to know everything by heart right now. However, in order to succeed in the class, you need to be able to read mathematical material at this level of abstraction and (lack of) detail.

Approaching 597

Let ๐ด be an infinite (not necessarily countable) set, and ๐‘“:๐ดโ†’โ„ a function. Suppose that for every integer ๐‘โ‰ฅ1 there exist finite subsets ๐ด๐‘+โŠ‚๐ด and ๐ด๐‘โˆ’โŠ‚๐ด such that:

  • (i) |๐‘“(๐›ผ)|โ‰ค๐‘โˆ’1 for all ๐›ผโˆˆ๐ด\(๐ด๐‘+โˆช๐ด๐‘โˆ’);

  • (ii) โˆ‘๐›ผโˆˆ๐ด๐‘+๐‘“(๐›ผ)โ‰ฅ๐‘;

  • (iii) โˆ‘๐›ผโˆˆ๐ด๐‘โˆ’๐‘“(๐›ผ)โ‰คโˆ’๐‘.

Prove that for any ๐‘โ‰ฅ1, there exists a finite subset ๐ต๐‘โŠ‚๐ด such that

|597โˆ’โˆ‘๐›ผโˆˆ๐ต๐‘๐‘“(๐›ผ)|โ‰ค1๐‘.
Proof

We first take ๐ด๐‘=๐ด๐‘+โ‹ƒ๐ด๐‘โˆ’โŠ‚๐ด s.t. |๐‘“(๐›ผ)|โ‰ค๐‘โˆ’1 for all ๐›ผโˆˆ๐ด\๐ด๐‘, as given by the conditions.
Now we define ๐‘๐‘œ๐‘ (๐ด๐‘)โ‰”{๐›ผโˆˆ๐ด๐‘โˆฃ๐‘“(๐›ผ)โ‰ฅ0} and ๐‘›๐‘’๐‘”(๐ด๐‘)โ‰”{๐›ผโˆˆ๐ด๐‘โˆฃ๐‘“(๐›ผ)<0}.
Let ๐‘”๐‘Ž๐‘โ‰”โˆ‘๐›ผโˆˆ๐ด๐‘๐‘“(๐›ผ)โˆ’597. This is a real number since ๐ด๐‘ is finite.

Case 1: if ๐‘”๐‘Ž๐‘<0, then we need to fill in more elements whose image under ๐‘“ sum up to be positive to make the sum close to 597 from below.
We then take a finite set ๐ต๐‘+โŠ‚๐ด s.t. โˆ‘๐›ผโˆˆ๐ต๐‘+๐‘“(๐›ผ)โ‰ฅโŒˆโˆ’๐‘”๐‘Ž๐‘+โˆ‘๐›ผโˆˆ๐‘๐‘œ๐‘ (๐ด๐‘)๐‘“(๐›ผ)โŒ‰.
Since ๐ต๐‘+โ‹‚๐ด๐‘โŠ‚๐ด๐‘, we have

โˆ‘๐›ผโˆˆ๐ต๐‘+โ‹‚๐ด๐‘๐‘“(๐›ผ)โ‰คโˆ‘๐›ผโˆˆ๐‘๐‘œ๐‘ (๐ด๐‘)๐‘“(๐›ผ)

and since ๐ต๐‘+=(๐ต๐‘+\๐ด๐‘)โˆ(๐ต๐‘+โ‹‚๐ด๐‘), we have

โˆ‘๐›ผโˆˆ๐ต๐‘+๐‘“(๐›ผ)=โˆ‘๐›ผโˆˆ๐ต๐‘+\๐ด๐‘๐‘“(๐›ผ)+โˆ‘๐›ผโˆˆ๐ต๐‘+โ‹‚๐ด๐‘๐‘“(๐›ผ)

By (1) and (2), it is clear that

โˆ‘๐›ผโˆˆ๐ต๐‘+\๐ด๐‘๐‘“(๐›ผ)โ‰ฅโˆ’๐‘”๐‘Ž๐‘

(3) means that the elements in ๐ต๐‘+\๐ด๐‘ have big enough image sum to fill the gap. And by definition, for all ๐›ผโˆˆ๐ต๐‘+\๐ด๐‘, we have |๐‘“(๐›ผ)|โ‰ค1๐‘. This means that each element in this finite ๐ต๐‘+\๐ด๐‘ takes up only a small portion of the sum, bounded by 1/๐‘. Together with (3), it follows that there is some subset ๐ต๐‘โ€ฒโŠ‚๐ต๐‘+\๐ด๐‘ s.t. โˆ‘๐›ผโˆˆ๐ต๐‘โ€ฒ๐‘“(๐›ผ)โˆˆ[โˆ’๐‘”๐‘Ž๐‘โˆ’1/๐‘,โˆ’๐‘”๐‘Ž๐‘+1/๐‘]. So for the finite set ๐ด๐‘โ‹ƒ๐ต๐‘โ€ฒ, we have

โˆ‘๐›ผโˆˆ๐ด๐‘โ‹ƒ๐ต๐‘โ€ฒ๐‘“(๐›ผ)=โˆ‘๐›ผโˆˆ๐ด๐‘๐‘“(๐›ผ)+โˆ‘๐›ผโˆˆ๐ต๐‘โ€ฒ๐‘“(๐›ผ)โˆˆ[597โˆ’1/๐‘,597+1/๐‘]
Figureย 1:

Case 2: if ๐‘”๐‘Ž๐‘>0, then we need to fill in more elements whose image under ๐‘“ sum up to be negative to make the sum close to 597 from above.
We then take finite ๐ต๐‘โˆ’โŠ‚๐ด s.t. โˆ‘๐›ผโˆˆ๐ต๐‘โˆ’๐‘“(๐›ผ)โ‰คโŒŠโˆ’๐‘”๐‘Ž๐‘+โˆ‘๐›ผโˆˆ๐‘›๐‘’๐‘”(๐ด๐‘)๐‘“(๐›ผ)โŒ‰.
For the same reason as case 1, we get

โˆ‘๐›ผโˆˆ๐ต๐‘โˆ’\๐ด๐‘๐‘“(๐›ผ)โ‰คโˆ’๐‘”๐‘Ž๐‘

And by definition, for all ๐›ผโˆˆ๐ต๐‘โˆ’\๐ด๐‘, we have |๐‘“(๐›ผ)|โ‰ค1๐‘. Together with (5), it follows that there is some subset ๐ต๐‘โ€ฒโŠ‚๐ต๐‘โˆ’\๐ด๐‘ s.t. โˆ‘๐›ผโˆˆ๐ต๐‘โ€ฒ๐‘“(๐›ผ)โˆˆ[โˆ’๐‘”๐‘Ž๐‘โˆ’1/๐‘,โˆ’๐‘”๐‘Ž๐‘+1/๐‘]. So for the finite set ๐ด๐‘โ‹ƒ๐ต๐‘โ€ฒ, we have

โˆ‘๐›ผโˆˆ๐ด๐‘โ‹ƒ๐ต๐‘โ€ฒ๐‘“(๐›ผ)=โˆ‘๐›ผโˆˆ๐ด๐‘๐‘“(๐›ผ)+โˆ‘๐›ผโˆˆ๐ต๐‘โ€ฒ๐‘“(๐›ผ)โˆˆ[597โˆ’1/๐‘,597+1/๐‘]


Case 3: ๐‘”๐‘Ž๐‘=0, then we are done.
This finishes the proof of the statement.

โ–ก

Limsup and Liminf

Let ๐‘‹ be a nonempty set, and ๐ด,๐ต subsets of ๐‘‹. Define a sequence (๐ธ๐‘›)๐‘›=1โˆž of subsets of ๐‘‹ by

๐ธ๐‘›={๐ดif ๐‘› is a prime number,๐ตotherwise.

Characterize the sets limโ€‰sup๐ธ๐‘› and limโ€‰inf๐ธ๐‘› (see ยง0.1 in Folland for notation).

Solution

By definition,

limโ€‰sup(๐ธ๐‘›)=โ‹‚๐‘˜=1โˆžโ‹ƒ๐‘›=๐‘˜โˆž๐ธ๐‘›

For each ๐‘˜โˆˆโ„•, there are infinitely many ๐‘›โ‰ฅ๐‘˜ such that ๐‘› is prime, and also there are infinitely many ๐‘›โ‰ฅ๐‘˜ such that ๐‘› is not prime. So โ‹ƒ๐‘›=๐‘˜โˆž๐ธ๐‘›=๐ดโ‹ƒ๐ต. Therefore

limโ€‰sup(๐ธ๐‘›)=โ‹‚๐‘˜=1โˆž(๐ดโ‹ƒ๐ต)=๐ดโ‹ƒ๐ต

By definition,

limโ€‰inf(๐ธ๐‘›)=โ‹ƒ๐‘˜=1โˆžโ‹‚๐‘›=๐‘˜โˆž๐ธ๐‘›

For each ๐‘˜โˆˆโ„•, there are infinitely many ๐‘›โ‰ฅ๐‘˜ such that ๐‘› is prime, and also there are infinitely many ๐‘›โ‰ฅ๐‘˜ such that ๐‘› is not prime. So โ‹‚๐‘›=๐‘˜โˆž๐ธ๐‘›=๐ดโ‹‚๐ต. Therefore

limโ€‰inf(๐ธ๐‘›)=โ‹ƒ๐‘˜=1โˆž(๐ดโ‹‚๐ต)=๐ดโ‹‚๐ต

Polynomial Convergence

Let ๐‘“:โ„คโ‰ฅ0ร—โ„คโ‰ฅ0โ†’โ„ be a function with the property that for every polynomial

๐‘(๐‘ฅ)=๐‘ฅ๐‘‘+๐‘Ž1๐‘ฅ๐‘‘โˆ’1+โ‹ฏ+๐‘Ž๐‘‘

with integer coefficients, we have that

lim๐‘›โ†’โˆž๐‘“(๐‘›,๐‘(๐‘›))=lim๐‘›โ†’โˆž๐‘“(๐‘(๐‘›),๐‘›)=0.

Does it follow that ๐‘“(๐‘š,๐‘›)โ†’0 as ๐‘š,๐‘›โ†’โˆž? In other words, given ๐œ–>0, does there exist ๐‘โ‰ฅ0 such that |๐‘“(๐‘š,๐‘›)|<๐œ– whenever |๐‘š|,|๐‘›|โ‰ฅ๐‘? Give a proof or a counterexample.

Solution

Consider this function:

๐‘“(๐‘š,๐‘›)={1,if ๐‘š=2๐‘›0,otherwise

Let ๐‘ be arbitrary polynomial with integer coefficients. Then there must be at most finite ๐‘› such that ๐‘(๐‘›)=2๐‘›. This is guaranteed by the asymptotic behavior of polynomial and exponential function: lim๐‘›โ†’โˆž๐‘(๐‘›)2๐‘›=0. So there exists some ๐‘โˆˆโ„• s.t. ๐‘(๐‘›)2๐‘›<1/2 for all ๐‘›โ‰ฅ๐‘, therefore ๐‘“(๐‘(๐‘›),๐‘›) is eventually 0.
Also, there must be at most finite ๐‘› such that 2๐‘(๐‘›)=๐‘›, i.e. ๐‘(๐‘›)=log2๐‘›. This is guaranteed by the asymptotic behavior of polynomial and logarithmic function: lim๐‘›โ†’โˆžlog2๐‘›๐‘(๐‘›)=0. So there exists some ๐‘โˆˆโ„• s.t. log2๐‘›๐‘(๐‘›)<1/2 for all ๐‘›โ‰ฅ๐‘, therefore ๐‘“(๐‘›,๐‘(๐‘›)) is eventually 0.
This confirms that lim๐‘›โ†’โˆž๐‘“(๐‘›,๐‘(๐‘›))=lim๐‘›โ†’โˆž๐‘“(๐‘(๐‘›),๐‘›)=0 for any polynomial ๐‘ with integer coefficients.
Then we consider the sequence ((2๐‘›,๐‘›))๐‘›โˆˆโ„•. For any ๐‘›โˆˆโ„•, ๐‘“((2๐‘›,๐‘›))=1, so the sequential limit is 1. This completes the counterexample.

Figureย 2:

1 Problem solving

Recall: Given mspace (๐‘‹,๐’œ๏ธ€,๐œ‡) ไปฅๅŠ ๐‘“:๐‘‹โ†’โ„‚ mble, ๆˆ‘ไปฌๅฏไปฅ define distribution function:

๐œ†๐‘“:(0,โˆž)โ†’[0,โˆž]

by

๐œ†๐‘“(๐›ผ)=๐œ‡({|๐‘“|>๐›ผ})

Chebyshevs ineq:

๐œ†๐‘“(๐›ผ)โ‰ค(โˆฅ๐‘“โˆฅ๐‘๐›ผ)๐‘

for 0<๐‘<โˆž.
Today: Problem Solving

Proposition 1.1

ๅฏนไบŽไปปๆ„ 0<๐‘<โˆž, ๆˆ‘ไปฌๆœ‰:

โˆซ๐‘‹|๐‘“|๐‘๐‘‘๐œ‡=โˆซ0โˆž๐‘๐›ผ๐‘โˆ’1๐œ†๐‘“(๐›ผ)๐‘‘๐›ผ

ๅทฆ่พนๆ˜ฏ integral on ๐‘‹, ๅณ่พนๆ˜ฏ integral on โ„.

Proof

Sketch: Step 1: ๐‘“ simple โŸน |๐‘“| simple.

โ–ก

Write

|๐‘“|=โˆ‘๐‘—=1๐‘๐‘๐‘—๐œ’๐ด๐‘—

where ๐ด๐‘— disjoint, ๐‘1>๐‘2>โ‹ฏ>๐‘๐‘>0 This implies:

โˆซ|๐‘“|๐‘๐‘‘๐œ‡=โˆ‘๐‘—=1๐‘๐‘๐‘—๐‘๐‘Ÿ๐‘—,๐‘Ÿ๐‘—=๐œ‡(๐ด๐‘—)

Then

๐œ†๐‘“(๐›ผ)={โˆ‘๐‘—=1๐‘๐‘Ÿ๐‘—,0<๐›ผ<๐‘๐‘โˆ‘๐‘—=1๐‘›โˆ’1๐‘Ÿ๐‘—,๐‘๐‘›โ‰ค๐›ผ<๐‘๐‘›โˆ’1,2โ‰ค๐‘›โ‰ค๐‘0,๐›ผโ‰ฅ๐‘1

ไปŽ่€Œ

โˆซ0โˆž๐‘๐›ผ๐‘โˆ’1๐œ†๐‘“(๐›ผ)๐‘‘๐›ผ=(โˆ‘๐‘—=1๐‘๐‘Ÿ๐‘—)โˆซ0๐‘๐‘๐‘๐›ผ๐‘โˆ’1๐‘‘๐›ผ+โˆ‘๐‘›=2๐‘(โˆ‘๐‘—=1๐‘›โˆ’1๐‘Ÿ๐‘—)โˆซ๐‘๐‘›๐‘๐‘›โˆ’1๐‘๐›ผ๐‘โˆ’1๐‘‘๐›ผ=

Step 2: ๐‘“ general.
Use: โˆƒ simple functions ๐‘”๐‘›โ‰ฅ0 s.t. ๐‘”๐‘›โ†—๏ธŽ|๐‘“|.
MCT โŸน

โˆซ๐‘‹|๐‘“|๐‘๐‘‘๐œ‡=lim๐‘›โ†’โˆžโˆซ๐‘‹๐‘”๐‘›๐‘๐‘‘๐œ‡

Also,

๐œ†๐‘”๐‘›โ†—๏ธŽCFB๐œ†๐‘“pointwisely on(0,โˆž)

ไปŽ่€Œ MCT โŸน

lim๐‘›โ†’โˆžโˆซ0โˆž๐‘๐›ผ๐‘โˆ’1๐œ†๐‘”๐‘›(๐›ผ)๐‘‘๐›ผโ†’โˆซ0โˆž๐‘๐›ผ๐‘โˆ’1๐œ†๐‘“(๐›ผ)๐‘‘๐›ผ

๐œ†๐‘“(๐›ผ)=๐œ‡({|๐‘“|>๐›ผ}), ไปฅๅŠ {|๐‘“|>๐›ผ}=โ‹ƒ1โˆž{๐‘”๐‘›>๐›ผ} increasing union.

Example 1.1

Let ๐‘“:[0,1]โ†’โ„ be abs ctn. Suppose ๐‘“(0)=0 ไปฅๅŠ ๐‘“1โˆˆ๐ฟ2([0,1]).
Show that the limit

lim๐‘ฅโ†’0+๐‘ฅโˆ’1/2๐‘“(๐‘ฅ)

exists, ๅนถ compute it.
What could the limit be? Must be 0.

Solution

Use FTOC, can recover ๐‘“ from ๐‘“โ€ฒ.

๐‘“(๐‘ฅ)=๐‘“(0)+โˆซ0๐‘ฅ๐‘“โ€ฒ(๐‘ก)๐‘‘๐‘ก,0โ‰ค๐‘ฅโ‰ค1

ไฝฟ็”จ Hรถlder with ๐‘=๐‘ž=2 (Cauchy-Swartz):

|๐‘“(๐‘ฅ)|โ‰คโˆซ0๐‘ฅ|๐‘“โ€ฒ(๐‘ก)|๐‘‘๐‘ก=โˆซ0๐‘ฅ|๐‘“โ€ฒ(๐‘ก)|1๐‘‘๐‘กโ‰ค(โˆซ0๐‘ฅ|๐‘“โ€ฒ(๐‘ก)|2)12๐‘ฅ12

ไปŽ่€Œ

๐‘ฅโˆ’1/2|๐‘“(๐‘ฅ)|โ‰คโˆซ0๐‘ฅ|๐‘“โ€ฒ(๐‘ก)|2๐‘‘๐‘ก

Use fact: ๐‘”=๐ฟ1(๐‘‹,๐’œ๏ธ€,๐œ‡)โŸนโˆ€๐œ–>0,โˆƒ๐›ฟ>0 s.t. for all ๐œ‡(๐ธ)<๐›ฟ we have โˆซ๐ธ|๐‘”|๐‘‘๐œ‡<๐œ–.
(Proof of this fact: use approx by simple functions ๅฏๅพ—).
็„ถๅŽ use approx by simple functions, apply to ๐‘”=|๐‘“โ€ฒ|2, ๐œ‡=๐‘š, ๐ธ=[0,๐‘ฅ], ไบŽๆ˜ฏๅพ—ๅˆฐ

โˆซ0๐‘ฅ|๐‘“โ€ฒ(๐‘ก)|2๐‘‘๐‘กโ†’๐‘ฅโ†’00
Example 1.2

Let ๐‘“:โ„๐‘›โ†’โ„ be a function.
Assume: ๅฏนไบŽ โˆ€๐œ–>0, ้ƒฝๅญ˜ๅœจ Lebesgue mble functions ๐‘”,โ„Žโˆˆ๐ฟ1(๐‘š) s.t.

๐‘”(๐‘ฅ)โ‰ค๐‘“(๐‘ฅ)โ‰คโ„Ž(๐‘ฅ)โˆ€๐‘ฅโˆˆโ„๐‘›

ๅนถไธ”

โˆซโ„๐‘›(โ„Žโˆ’๐‘”)๐‘‘๐‘š<๐œ–

Prove that: ๐‘“ ไนŸๆ˜ฏ Lebesgue mble ็š„, ๅนถไธ” ๐‘“โˆˆ๐ฟ1(๐‘š).

Proof

By assumption: Given ๐‘˜โˆˆโ„•, ๅญ˜ๅœจ ๐‘”๐‘˜,โ„Ž๐‘˜โˆˆ๐ฟ1(โ„๐‘›) s.t.

๐‘”๐‘˜โ‰ค๐‘“โ‰คโ„Ž๐‘˜,โˆซ(โ„Ž๐‘˜โˆ’๐‘”๐‘˜)<1๐‘˜

Idea: ๐‘“=limโ€‰sup๐‘”๐‘˜=limโ€‰infโ„Ž๐‘˜ ?
ๆˆ‘ไปฌๅบ”่ฏฅ try to prove: for a.e. ๐‘ฅ ้ƒฝๆœ‰ 0โ‰คโ„Ž๐‘˜(๐‘ฅ)โˆ’๐‘”๐‘˜(๐‘ฅ)โ†’0.
Use Fatouโ€™s Lemma:

โˆซlimโ€‰inf๐‘˜โ†’โˆž(โ„Ž๐‘˜โˆ’๐‘”๐‘˜)โ‰คlimโ€‰inf๐‘˜โ†’โˆžโˆซ(โ„Ž๐‘˜โˆ’๐‘”๐‘˜)=0

่€Œ โ„Ž๐‘˜โˆ’๐‘”๐‘˜โ‰ฅ0, ๅ› ่€Œ This means:

limโ€‰inf๐‘˜โ†’โˆž(โ„Ž๐‘˜โˆ’๐‘”๐‘˜)=0for a.e. ๐‘ฅ

ไธ”ๆˆ‘ไปฌ็Ÿฅ้“

limโ€‰inf๐‘˜โ†’โˆž(โ„Ž๐‘˜โˆ’๐‘“)โ‰คlimโ€‰inf๐‘˜โ†’โˆž(โ„Ž๐‘˜โˆ’๐‘”๐‘˜)=0for a.e. ๐‘ฅ

ไปŽ่€Œ

๐‘“(๐‘ฅ)=limโ€‰inf๐‘˜โ†’โˆžโ„Ž๐‘˜(๐‘ฅ)for a.e. ๐‘ฅ

This proves that, ๐‘“ is Lebesgue measurable.

โ–ก

Example 1.3

Prove that:

lim๐‘›โ†’โˆžโˆซ๐ธsin(๐‘›๐‘ฅ)๐‘‘๐‘ฅ=0

for every bounded Borel set ๐ธโŠ‚โ„.

Proof

Step 1: ๐ธ=(๐‘Ž,๐‘) ๆ˜ฏไธ€ไธช interval.

โˆซ๐ธsin(๐‘›๐‘ฅ)๐‘‘๐‘ฅ=[โˆ’1๐‘›cos(๐‘›๐‘ฅ)]๐‘Ž๐‘

ไปŽ่€Œ

|โˆซ๐ธsin(๐‘›๐‘ฅ)๐‘‘๐‘ฅ|โ‰ค2๐‘›โ†’๐‘›โ†’โˆž0

Step 2: ๐ธ ๆ˜ฏไธ€ไธช finite union of disjoint open intervals.
Same as Step 1.
Step 3: General Case.
Fix ๐œ–>0.
Then by outer regularity: ๅญ˜ๅœจ some ๐‘ˆ ไธบ finite disjoint union of open intervals, ไฝฟๅพ—

๐‘š(๐‘ˆฮ”๐ธ)<๐œ–

ไปŽ่€Œ

|โˆซ๐ธ๐‘“๐‘›=โˆซ๐‘ˆ๐‘“๐‘›|<|โˆซ๐‘ˆฮ”๐ธ๐‘“๐‘›|โ‰ค๐‘š(๐‘ˆฮ”๐ธ)<๐œ–

ๅ› ่€Œ

|โˆซ๐ธ๐‘“๐‘›|<|โˆซ๐‘ˆ๐‘“๐‘›|+๐œ–

for all ๐‘›. ๅนถไธ” By step 2:

limโ€‰sup๐‘›โ†’โˆž|โˆซ๐‘ˆ๐‘“๐‘›|+๐œ–=0+๐œ–

ๅ› ่€Œ

limโ€‰sup๐‘›โ†’โˆž|โˆซ๐ธ๐‘“๐‘›|โ‰ค๐œ–

Since ๐œ– arbitrary, ๅพ—่ฏ.

โ–ก

Example 1.4

Let ๐ธโŠ‚โ„ be a Borel set, with ๐‘š(๐ธ)>0.
Set ๐‘“:โ„โ†’โ„ be mble, nonneg, ๅนถไธ” โˆซ๐‘“>0.
Prove that: ๅญ˜ๅœจ ๐‘กโˆˆโ„ s.t.

โˆซ๐ธ+๐‘ก๐‘“>0
Proof

Claim 1: STS to assume ๐‘“ simple.
Proof of Claim 1: ๅฏนไบŽ ๐‘“, can find seq of simple functions 0โ‰ค๐‘“๐‘›โ‰ค๐‘“, s.t. ๐‘“๐‘›โ†—๏ธŽ๐‘“.
By MCT,

โ–ก

2 problem solving-III

Example 2.5

Let ๐‘“โˆˆ๐ฟ๐‘™๐‘œ๐‘2(โ„).
Assume

โˆซ๐‘Ž๐‘Ž|๐‘ก||๐‘“(๐‘ฅ+๐‘ก)|๐‘‘๐‘กโ‰ฅ23๐‘Ž2

for all ๐‘Ž>0, ๐‘ฅโˆˆโ„.
Now show: |๐‘“(๐‘ฅ)|โ‰ฅ1 for a.e. ๐‘ฅ.

Proof

WLOG ๅฏไปฅๅ‡่ฎพ ๐‘“ ๆ˜ฏ nonneg ็š„. (๐‘“โ†ฆ|๐‘“|).
WTS: |๐‘“(๐‘ฅ)|โ‰ฅ1 for a.e. ๐‘ฅ.
Claim 1: by LDT, it STS:

12๐‘Žโˆซโˆ’๐‘Ž๐‘Ž๐‘“(๐‘ฅ+๐‘ก)๐‘‘๐‘กโ‰ฅ1

for all ๐‘ฅโˆˆโ„.
ๆˆ‘ไปฌ try Cauchy Swartz:

โˆซโˆ’๐‘Ž๐‘Ž|๐‘ก|๐‘“(๐‘ฅ+๐‘ก)๐‘‘๐‘กโ‰ค(โˆซโˆ’๐‘Ž๐‘Ž๐‘ก2๐‘‘๐‘ก)1/2(โˆซโˆ’๐‘Ž๐‘Ž๐‘“(๐‘ฅ+๐‘ก)2๐‘‘๐‘ก)1/2

ๆˆ‘ไปฌ็Ÿฅ้“: ๅทฆ่พน โ‰ฅ23๐‘Ž2, ่€Œๅณ่พน็ฌฌไธ€้กน (โˆซโˆ’๐‘Ž๐‘Ž๐‘ก2๐‘‘๐‘ก)1/2 ๆ˜ฏๅฏไปฅ่ฎก็ฎ—็š„: ็ญ‰ไบŽ (2๐‘Ž33)1/2.
ไบŽๆ˜ฏ, ๆˆ‘ไปฌๅพ—ๅˆฐ

โˆซโˆ’๐‘Ž๐‘Ž๐‘“(๐‘ฅ+๐‘ก)2๐‘‘๐‘กโ‰ฅ2๐‘Ž

ไปŽ่€Œ:

12๐‘Žโˆซโˆ’๐‘Ž๐‘Ž๐‘“(๐‘ฅ+๐‘ก)2๐‘‘๐‘กโ‰ฅ1

็„ถๅŽ by LDT:

12๐‘Žโˆซโˆ’๐‘Ž๐‘Ž๐‘“(๐‘ฅ+๐‘ก)2๐‘‘๐‘ก=12๐‘Žโˆซ๐‘ฅโˆ’๐‘Ž๐‘ฅ+๐‘Ž๐‘“(๐‘ฆ)2๐‘‘๐‘ฆ=๐‘“(๐‘ฅ)2

for a.e. ๐‘ฅ. ๅ› ่€Œ

๐‘“(๐‘ฅ)2โ‰ฅ1for a.e. ๐‘ฅ

ไบŽๆ˜ฏ

|๐‘“(๐‘ฅ)|โ‰ฅ1for a.e. ๐‘ฅ

โ–ก

Example 2.6

Prove or disprove: ๅฏนไบŽ bounded open set ๐ธโŠ‚โ„, ๅฎƒ็š„ boundary ๆ˜ฏๅฆไธ€ๅฎšๆปก่ถณ ๐‘š(๐œ•๐ธ)=0 ?

Solution

Astonishingly ่ฟ™ไธช้—ฎ้ข˜็š„ๅ›ž็ญ”ๆ˜ฏๅฆๅฎš็š„. ๆˆ‘ไปฌๅฏไปฅๆž„้€ 

3 extra topics

3.1 Minkowski ineq for integral

3.2 convolution

ๆˆ‘ไปฌๅทฒ็ป่ฏๆ˜Žไบ†, for 1โ‰ค๐‘<โˆž,

๐ถ๐‘0(โ„๐‘›)โŠ‚๐ฟ๐‘(โ„๐‘›)dense subset

What about for ๐‘=โˆž? ็ญ”ๆกˆไนŸๆ˜ฏ true ็š„, ๆˆ‘ไปฌ้œ€่ฆ็”จๅˆฐ convolution ๆฅ่ฏๆ˜Ž.

4 Use FTC and Tonelli for series

Let ๐‘”๐‘˜,๐‘˜=1,2,โ€ฆ, be a sequence of functions that are absolutely continuous on the interval [๐‘Ž,๐‘]. Suppose that there is a ๐‘โˆˆ[๐‘Ž,๐‘], such that the series โˆ‘๐‘˜=1โˆž๐‘”๐‘˜(๐‘) is convergent, and

โˆ‘๐‘˜=1โˆžโˆซ๐‘Ž๐‘|๐‘”๐‘˜โ€ฒ(๐‘ฅ)|๐‘‘๐‘ฅ<โˆž

(a) Show that โˆ‘๐‘˜=1โˆž๐‘”๐‘˜(๐‘ฅ) is convergent for all ๐‘ฅโˆˆ[๐‘Ž,๐‘]. (b) Let ๐‘“(๐‘ฅ)=โˆ‘๐‘˜=1โˆž๐‘”๐‘˜(๐‘ฅ). Show that ๐‘“ is absolutely continuous on [๐‘Ž,๐‘] and

๐‘“โ€ฒ(๐‘ฅ)=โˆ‘๐‘˜=1โˆž๐‘”๐‘˜โ€ฒ(๐‘ฅ) for almost every โˆˆ[๐‘Ž,๐‘]

โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

5 Use FTC and Holder

Let ๐‘“:[0,1]โ†’๐‘… be absolutely continuous, satisfy ๐‘“(0)=0 and ๐‘“โ€ฒโˆˆ๐ฟ2([0,1]). Show that

lim๐‘ฅโ†’0+๐‘ฅโˆ’1/2๐‘“(๐‘ฅ)

exists and determine the value of this limit. โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

6 Use density of compactly supported continuous functions in a suitable space

Let ๐‘“ be a real Lebesgue measurable function on the interval [0,1] such that โˆฅ๐‘“โˆฅโˆž<โˆž. Show that for any ๐œ€,๐›ฟ>0, there is a continuous function ๐‘” on [0,1] such that ๐‘š{๐‘ฅโˆˆ[0,1]:|๐‘“(๐‘ฅ)โˆ’๐‘”(๐‘ฅ)|>๐œ€}<๐›ฟ. โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

7 Use one of the convergence theorems

Let A be a sequence of measurable subsets of [0,1] such that inf๐‘š(๐ด๐‘›)>0, where ๐‘š stands for the Lebesgue measure. (a) Prove that there exists ๐‘ฅโˆˆ[0,1] which belongs to infinitely many of the sets ๐ด๐‘›. (b) Does there necessarily exist a point which belongs to any of the sets ๐ด๐‘›, except finitely many? โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

8 How can we recover E from its indicator function

Let ๐ธโŠ‚โ„1. Show that the characteristic function ๐œ’๐ธ(๐‘ฅ) is the limit of a sequence of continuous functions if and only if ๐ธ is both ๐น๐œŽ and ๐บ๐›ฟ.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

9 be an artisan

Let ๐‘“:[0,1]โ†’โ„ be a positive function of bounded variation. (a) Show that if inf(๐‘“)>0, then the function ๐‘”(๐‘ฅ)=1/๐‘“(๐‘ฅ) is also of bounded variation on [0,1]. (b) Give an example of a positive function ๐‘“:[0,1]โ†’โ„ of bounded variation such that ๐‘”(๐‘ฅ)=1/๐‘“(๐‘ฅ) is integrable but not of bounded variation.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

10 Use a suitable theorem allowing you to differentiate exp(๐‘”) under the integral sign

Let ๐‘“ be a real Lebesgue measurable function on the interval [0,1] such that โˆฅ๐‘“โˆฅโˆž<โˆž. For ๐›ผโˆˆโ„ define a function ๐‘”(๐›ผ) by

๐‘”(๐›ผ)=log[โˆซ01exp[๐›ผ๐‘“(๐‘ฅ)]๐‘‘๐‘ฅ]

โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

(a) Prove that the function ๐‘”(โ‹…) is twice continuously differentiable and that ๐‘”โ€ณ(๐›ผ)โ‰ฅ0 for all ๐›ผโˆˆโ„, i.e. the function ๐‘”(โ‹…) is convex. (b) Prove that if ๐‘“ is a non-constant function, i.e. ๐‘š{๐‘ฅโˆˆ[0,1]:|๐‘“(๐‘ฅ)โˆ’๐‘|โ‰ 0}>0 for all constants ๐‘โˆˆโ„, then ๐‘”โ€ณ(๐›ผ)>0,๐›ผโˆˆโ„.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

11 Use DCT

Let

๐‘“โˆˆ๐ฟ1([0,1],๐‘‘๐‘ฅ)

Find:

lim๐‘›โ†’โˆž1๐‘›โˆซ01log(1+๐‘’๐‘›๐‘“(๐‘ฅ))๐‘‘๐‘ฅ

โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

12 Use Egoroff and Hรถlder

Let {๐‘“๐‘›} be a sequence of functions in ๐ฟ๐‘(โ„๐‘›),1<๐‘<โˆž, which converge almost everywhere to a function ๐‘“โˆˆ๐ฟ๐‘(โ„๐‘›), and suppose that there is a constant ๐‘€ such that โˆฅ๐‘“๐‘›โˆฅ๐‘โ‰ค๐‘€ for all ๐‘›. Show that for every ๐‘”โˆˆ๐ฟ๐‘ž(โ„๐‘›),๐‘ž the conjugate of ๐‘,

โˆซ๐‘“๐‘”=lim๐‘›โ†’โˆžโˆซ๐‘“๐‘›๐‘”

Is the statement true for ๐‘=1 ? (Hint: you may want to use Egorovโ€™s Theorem.)โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

13 Read up on HL

Let ๐‘“(โ‹…) be a locally integrable function on โ„๐‘› and ๐‘€๐‘“ the corresponding Hardy-Littlewood maximal function

๐‘€๐‘“(๐‘ฅ)=sup๐‘…>01|๐ต(๐‘ฅ,๐‘…)|โˆซ๐ต(๐‘ฅ,๐‘…)|๐‘“(๐‘ฆ)|๐‘‘๐‘ฆ,๐‘ฅโˆˆโ„๐‘›

where ๐ต(๐‘ฅ,๐‘…) denotes the ball centered at ๐‘ฅ with radius ๐‘…. a) Show that if ๐‘“ is integrable on โ„๐‘› then sup๐œ†>0๐œ†๐‘š{๐‘ฅโˆˆโ„๐‘›:|๐‘“(๐‘ฅ)|>๐œ†}<โˆž. b) Let ๐‘“ be the function

๐‘“(๐‘ฅ)={1 if |๐‘ฅ|<10 if |๐‘ฅ|โ‰ฅ1

Show that ๐‘€๐‘“ is not integrable on โ„๐‘›, but sup๐œ†>0๐œ†๐‘š{๐‘ฅโˆˆโ„๐‘›:๐‘€๐‘“(๐‘ฅ)>๐œ†}< โˆž.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

14 Use density of such functions g somewhere, and then Hรถlder.

Fix 1<๐‘<โˆž. Let ๐‘“โˆˆ๐ฟ๐‘(๐ธ), where ๐ธ is a measurable subset of โ„๐‘‘. Assume that

โˆซ๐ธ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)๐‘‘๐‘ฅ=0

for all compactly supported continuous functions ๐‘”:โ„๐‘‘โ†’โ„. Is ๐‘“(๐‘ฅ)=0 for almost every ๐‘ฅ in ๐ธ ? If your answer is positive, prove it. Otherwise, given a counterexample.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

15 Fubini and Tonelli

Suppose that ๐‘“(๐‘ฅ),๐‘ฅ>0, is a real valued Lebesgue measurable square integrable function. (a) Prove that for any ๐›ผ>0, the inequality 2|๐‘“(๐‘ง)||๐‘“(๐‘ฆ)|โ‰ค๐›ผ๐‘“(๐‘ง)2+๐‘“(๐‘ฆ)2/๐›ผ holds for all ๐‘ง,๐‘ฆ,๐›ผ>0. (b) Express the double integral

โˆซ0โˆžโˆซ0โˆž|๐‘“(๐‘ง)||๐‘“(๐‘ฆ)|๐‘ฆ+๐‘ง๐‘‘๐‘ง๐‘‘๐‘ฆ

as an integral over the region {0<๐‘ง<๐‘ฆ<โˆž}. (c) Show using your work from (a) and (b) that |๐‘“(๐‘ง)||๐‘“(๐‘ฆ)|/(๐‘ฆ+๐‘ง),๐‘ฆ,๐‘ง>0, is integrable and

โˆซ0โˆžโˆซ0โˆž|๐‘“(๐‘ง)||๐‘“(๐‘ฆ)|๐‘ฆ+๐‘ง๐‘‘๐‘ง๐‘‘๐‘ฆโ‰ค4โˆซ0โˆž๐‘“(๐‘ฅ)2๐‘‘๐‘ฅ

Hint: Use the inequality in (a) with ๐›ผ=(๐‘ง/๐‘ฆ)1/2.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

16 Try a very nice function f first

Let {๐‘“๐‘›(๐‘ฅ)} be a sequence of continuous, strictly positive functions on โ„ which converges uniformly to the function ๐‘“(๐‘ฅ). Suppose that all the functions {๐‘“๐‘›},๐‘“ are integrable. Is

lim๐‘›โ†’โˆžโˆซ๐‘“๐‘›(๐‘ฅ)๐‘‘๐‘ฅ=โˆซ๐‘“(๐‘ฅ)๐‘‘๐‘ฅ

Justify your answer.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

17 Use Lebesgue. Can you get the same equality for more sets E?

Let ๐‘“โˆˆ๐ฟ1([0,1],๐‘‘๐‘ฅ) be a function such that โˆซ๐ธ๐‘“(๐‘ฅ)๐‘‘๐‘ฅ=0 for any measurable set ๐ธโŠ‚[0,1] of Lebesgue measure .99. Prove that ๐‘“=0 a.e.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

18 Lebesgue

Let ๐‘“โˆˆ๐ฟ2(๐ผ), for any finite interval ๐ผโŠ‚โ„. Assume that

โˆซโˆ’๐‘Ž๐‘Ž|๐‘ก||๐‘“(๐‘ฅ+๐‘ก)|๐‘‘๐‘กโ‰ฅ23๐‘Ž2

for all ๐‘Ž>0 and ๐‘ฅโˆˆโ„. Show that |๐‘“(๐‘ฅ)|โ‰ฅ1 for a.e. ๐‘ฅโˆˆโ„.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

19 Integration can be a trick to prove that a nonnegative function canโ€™t be identically zero.

Let ๐‘“ and ๐‘” be nonnegative functions in ๐ฟ1(โ„). Suppose that each function is positive on some set of positive measure. (However, there need not be a single set of positive measure where both functions are positive.) Prove that the convolution

โ„Ž(๐‘ฅ)=โˆซโˆ’โˆžโˆž๐‘“(๐‘ฅโˆ’๐‘ก)๐‘”(๐‘ก)๐‘‘๐‘ก

is positive on some set of positive measure.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

20 Check what happens on some set {๐‘“<๐‘} with ๐‘<||๐‘“||โˆž

Let ๐ธ be a measurable subset of โ„ such that ๐‘š(๐ธ)<โˆž. Let ๐‘“โˆˆ๐ฟโˆž(๐ธ) with โˆฅ๐‘“โˆฅโˆž>0. Show that

lim๐‘›โ†’โˆžโˆฅ๐‘“โˆฅ๐‘›+1๐‘›+1โˆฅ๐‘“โˆฅ๐‘›๐‘›=โˆฅ๐‘“โˆฅโˆž

Here โˆฅ๐‘“โˆฅ๐‘›โ‰”โˆฅ๐‘“โˆฅ๐ฟ๐‘›(๐ธ),โˆฅ๐‘“โˆฅ๐‘›+1โ‰”โˆฅ๐‘“โˆฅ๐ฟ๐‘›+1(๐ธ).โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†

21 Use distribution functions

Let ๐‘“:โ„โ†’โ„ be a measurable function which has the property that

๐‘š(|๐‘“|>๐›ผ)โ‰ค11+๐›ผ3 for ๐›ผ>0

(a) Show that |๐‘“|๐‘ is integrable for ๐‘<3. (b) Give an example of a function satisfying the above for which |๐‘“|3 is not integrable.โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€† โ€†