344 CHAPTER 14. THE LEBESGUE INTEGRAL

Definition 14.12.4 Let E be any set in R. Then

m̄(E)≡ inf{m(F) : F is a Borel set and F ⊇ E} .

Proposition 14.12.5 The following hold.

1. m̄(F)≤ m̄(G) if F ⊆ G

2. If E is Borel, then m̄(E) = m(E)

3. m̄(∪∞

k=1Ek)≤ ∑k m̄(Ek)

4. If m̄(E) = 0, then there exists F Borel such that F ⊇ E and m(F) = 0.

5. For any E, there exists F Borel such that F ⊇ E and m̄(E) = m(F).

Proof: The first claim is obvious. Consider the second. By definition, m̄(E) ≤ m(E)because E ⊇ E. If F ⊇ E and F is Borel, then m(E) ≤ m(F) and so, taking the inf of allsuch F containing E, it follows that m(E)≤ m̄(E) . Now consider the third assertion.

If any Ek has m̄(Ek) = ∞, then there is nothing to show. Assume then that m̄(Ek)< ∞

for all k. Then by definition, there is Fk Borel, containing Ek such that m̄(Ek)+ε

2k ≥m(Fk) .Then m̄(∪kEk)≤

m̄(∪kFk) = m(∪kFk)≤∞

∑k=1

m(Fk)≤∑k

m̄(Ek)+ε

2k = ∑k

m̄(Ek)+ ε

Since ε is arbitrary, this shows 3.Finally consider 4.,5. Of course 4. is a special case of 5. If m̄(E) = ∞, let F = R.

Otherwise there exists Gn ⊇ E such that m(Gn) < m̄(E) + 1/2n and Gn is a Borel set.Then let Fn ≡ ∩n

k=1Gn. It follows that ∩nFn ≡ F is a Borel set containing E and m(F) =limn→∞ m(Fn) = 0.

Definition 14.12.6 Let | f |XB ∈ L1 (R,m) for every bounded Borel B. The HardyLittlewood maximal function M f (x) is defined as

M f (x)≡ sup1≥r>0

1m(B(0,r))

∫B(x,r)

| f |dm

You can try and show this function is measurable, but I will not do so here. The funda-mentally important inequality involving this maximal function is the following major resultwhich comes from the above Vitali covering theorem and is one of the main applicationsof the above Vitali covering theorem.

Theorem 14.12.7 Let | f | ∈ L1 (R,m). Then if λ > 0,

m̄([M f > λ ])≤ 5λ

∫| f |dm

Proof: If x ∈ [M f > λ ] , then there is rx < 1 such that

1m(B(0,rx))

∫B(x,rx)

| f |dm > λ , m(B(0,rx))≤1λ

∫B(x,rx)

| f |dm