9.1.3 The Sigma Algebra Of Lebesgue Measurable Sets

In the above section, Lebesgue measure was defined on a σ algebra ℱ^{n} which was the smallest
σ algebra which contains K, where K was the set of measurable rectangles, sets of the form
∏_{i=1}^{n}E_{i} for E_{i} a Lebesgue measurable set. However, this is not a complete measure space.
To see this, consider A×B where m_{1}

(B )

= 0 and A is a non measurable set. Then A×B is a
subset of ℝ ×B and m_{2}

(ℝ × B)

= 0. However, A×B cannot be in σ

(K)

because all sets E in
σ

(K )

have the property that E_{y}≡

{x : (x,y) ∈ E}

is a measurable set. In this case, pick
y ∈ B and

(A × B)

_{y} = A which is not measurable. This leads to the following
definition.

Definition 9.1.6The measure space for Lebesgue measure is

(ℝn,ℱn,mn )

where ℱ_{n}is the completion of

(ℝn,ℬ (ℝn ),mn)

where ℬ

(ℝn)

denotes the Borel sets isn dimensional measure defined on ℬ

(ℝn)

.

The important thing about Lebesgue measure is that the measure space is complete and it
is a regular measure space.

Theorem 9.1.7The measure space

(ℝn,ℱ ,m )
n n

is complete and regular. Thismeans that for every E ∈ ℱ_{n}, there exists an F_{σ}set F and a G_{δ}set G such thatG ⊇ E ⊇ F and m_{n}

(G ∖ F)

= 0. Also if f ≥ 0 is ℱ_{n}measurable, then there existsg ≤ f such that g is Borel measurable and g = f a.e.

Proof:Let E ∈ℱ_{n}. From Theorem 9.1.4 there exist Borel sets A,B such that
A ⊆ E ⊆ B and m_{n}

(B ∖A )

= 0. Now from Proposition 9.1.3, there exists a G_{δ} set G
containing B and an F_{σ} set F contained in A such that m_{n}

Now consider the last claim. Let s_{k} be an increasing sequence of simple functions which
converges pointwise to f. Say s_{k}

(x)

= ∑_{i=1}^{mk}c_{
i}X_{Ei}

(x)

where E_{i}∈ℱ_{n}. Then let F_{i}⊆ E_{i}
such that F_{i} is Borel measurable and m_{n}

(Ei ∖ Fi)

= 0. Letting N_{k} = ∪_{i=1}^{mk}

(Ei ∖ Fi)

, and
N^{′} = ∪_{k=1}^{∞}N_{k}, it follows m_{n}

(N′)

= 0. Now let N be a Borel measurable set such that N
has measure zero and N ⊇ N^{′}. Then each function in the sequence of simple functions given
by s_{k}^{′}≡X_{N}∑_{i=1}^{mk}c_{
i}X_{Fi}is Borel measurable, and the sequence converges to a Borel
measurable function which equals f off the exceptional set of measure zero N and equals 0 on
N.■

I defined

(ℱn,mn )

as the completion of

(ℬ(ℝn ),mn )

but the same thing would have been
obtained if I had used

(ℱn, mn)

. This can be shown from using the regularity of one
dimensional Lebesgue measure. You might try and show this.