232 CHAPTER 9. BASIC FUNCTION SPACES
This proves claim 1.Claim 2: If g ∈Cc (Rp), then
limr→0
1mp (B(x,r))
∫B(x,r)
|g(y)−g(x)|dmp (y) = 0
Proof: Since g is continuous at x, whenever r is small enough,
1mp (B(x,r))
∫B(x,r)
|g(y)−g(x)|dmp (y)≤1
mp (B(x,r))
∫B(x,r)
ε dmp (y) = ε.
This proves the claim.Now let g ∈Cc (Rp). Then from the above observations about continuous functions in
Claim 2,
mp
([x : limsup
r→0
1mp (B(x,r))
∫B(x,r)
| f (y)− f (x)|dmp (y)> ε
])(9.8)
≤ mp
([x : limsup
r→0
1mp (B(x,r))
∫B(x,r)
| f (y)−g(y)|dmp (y)>ε
2
])+mp
([x : limsup
r→0
1mp (B(x,r))
∫B(x,r)
|g(y)−g(x)|dmp (y)>ε
2
])+mp
([x : |g(x)− f (x)|> ε
2
]).
≤ mp
([M ( f −g)>
ε
2
])+mp
([| f −g|> ε
2
])(9.9)
Now∥ f −g∥1 ≥
∫[| f−g|> ε
2 ]| f −g|dmp ≥
ε
2mp
([| f −g|> ε
2
])and so using Claim 1 and 9.9, 9.8 is dominated by(
2ε+
5p
ε
)∫| f −g|dmp.
But by Theorem 9.4.2, g can be chosen to make the above as small as desired. Hence 9.8is 0.
mp
([limsup
r→0
1mp (B(x,r))
∫B(x,r)
| f (y)− f (x)|dmp (y)> 0])
≤∞
∑k=1
mp
([limsup
r→0
1mp (B(x,r))
∫B(x,r)
| f (y)− f (x)|dmp (y)>1k
])= 0
By completeness of mp this implies[limsup
r→0
1mp (B(x,r))
∫B(x,r)
| f (y)− f (x)|dmp (y)> 0]
is a set of mp measure zero. ■The following corollary is the main result referred to as the Lebesgue Differentiation
theorem.