410 CHAPTER 15. THE Lp SPACES

|| f −g||Lp(Rn,µ) <ε

3 . Let k be such that spt(g) ⊆ (−k,k)n . Now by the Weierstrass ap-proximation theorem there exists a polynomial q such that

||g−q||[−(k+1),k+1]n ≡ sup{|g(x)−q(x)| : x ∈ [−(k+1) ,(k+1)]n}

<ε

3µ ((−(k+1) ,k+1)n).

It follows

||g− τkq||[−(k+1),k+1]n = ||τkg− τkq||[−(k+1),k+1]n

<ε

3µ ((−(k+1) ,k+1)n).

Without loss of generality, it can be assumed this polynomial has all rational coefficients.Therefore, τkq ∈D .

||g− τkq||pLp(Rn)=

∫(−(k+1),k+1)n

|g(x)− τk (x)q(x)|p dµ

≤(

ε

3µ ((−(k+1) ,k+1)n)

)p

µ ((−(k+1) ,k+1)n)

<(

ε

3

)p.

It follows

|| f − τkq||Lp(Rn,µ) ≤ || f −g||Lp(Rn,µ)+ ||g− τkq||Lp(Rn,µ) <ε

3+

ε

3< ε.

This proves the theorem.

Corollary 15.3.3 Let Ω be any µ measurable subset of Rn and let µ be a Radon measure.Then Lp(Ω,µ) is separable. Here the σ algebra of measurable sets will consist of allintersections of measurable sets with Ω and the measure will be µ restricted to these sets.

Proof: Let D̃ be the restrictions of D to Ω. If f ∈ Lp(Ω), let F be the zero extensionof f to all of Rn. Let ε > 0 be given. By Theorem 15.3.1 or 15.3.2 there exists s ∈D suchthat ||F− s||p < ε . Thus

||s− f ||Lp(Ω,µ) ≤ ||s−F ||Lp(Rn,µ) < ε

and so the countable set D̃ is dense in Lp(Ω).

15.4 Continuity Of TranslationDefinition 15.4.1 Let f be a function defined on U ⊆ Rn and let w ∈ Rn. Then fw will bethe function defined on w+U by

fw(x) = f (x−w).