Page 365 - DMTH401_REAL ANALYSIS

Unit 30: Riemann's and Lebesgue

f(x h) f(x) Notes
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-
D f(x) = limsup [upper left derivative]
–
h  - h
0
f(x h) f(x)
+
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D f(x) = liminf [lower left derivative].
–
h  - h
0
-
+
f(x h) f(x) é f(x h) f(x)ù
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Note Here, limsup := lim ê sup ú , and similarly the others.
0
h  + h y + ë 0 h y h û
<
<
0
Example: Let f: (–1,1)   be f(0) = 0 and f(x) = x sin(1/x) for x = 0. Then, D f(0) = 1 = D f(0)
+
–
and D f(0) = –1 = D f(0) so that f is not differentiable at 0.
+ –
Notes That f is differentiable at x iff all the four Dini derivatives are equal and real
(i.e., different from ±¥). Since D f(x)  D f(x) and D f(x) < D f(x) by definition, we also see
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+
+ -
+
that f is differentiable at x iff the four Dini derivatives are real numbers satisfying D f(x) 
D f(x) and D f(x) ¥ D f(x).
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- +
[Lebesgue’s differentiation theorem] Let –¥  a < b  ¥, let f : (a, b)   be a monotone function
and let Y = {x  (a, b) : f is not differentiable at x}. Then m * L,1 [Y] = 0.
Proof: Since (a, b) can be written as a countable union of bounded open intervals, we may as well
assume (a, b) itself is bounded. Assume f is increasing and write µ*= m * L,1 . By the remark above,
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–
Y = Y  Y , where Y = {x  (a, b) : D + f(x) < D f(x)} and Y = {x  (a, b) : D f(x) < D f(x)}. We will
1 2 1 – 2 +
only show that µ*[Y ] = 0; the case of Y is similar.
1 2
Let  = {(r, s)   : r < s}, let X(r, s) = {x  (a,b): D f(x) < r < s < D f(x)} and note that Y = U X(r, s).
2
+
– 1 (rs)
Hence it suffices to show there µ*[X(r, s)] = 0 for every (r, s)  . Fix (r, s)  , write X = X(r, s) and
let  > 0 be arbitrary. Choose an open set U  (a, b) such that X  U and µ*[U] < µ*[X] + .
Since D f < r on X, for each x  X and  > 0 we can find a non-degenerate closed interval I(x, ) =
–
[x –, x]  U such that 0 <  <  and f(x) – f(x – ) < r. Then  = {I(x, ) : x  X,  > 0} is a Vitali cover
for X. By Vitali’s lemma, we can find finitely many pairwise disjoint intervals I ,…,I   such
1 k
that µ*[X\  k n 1 |I |] < .
=
n
Let V =  k n 1 int[I ]. Then, V is open, V  U, and µ*[X] –  < µ*[V]  µ*[U] < µ*[X] + . Let X = V  X.
=
n
+
Since D f > s on X, and hence on X, for each y  X and  > 0 we can find a non-degenerate closed
interval J(y,) = [y, y + ]  V (hence J(y,)  I for some n  {1,…,k}) such that 0 <  < , and
n
f(y + ) – f(y) > s. Then  = {J(y, ) : y  X,  > 0} is a Vitali cover for X. Again by Vitali’s lemma,
we can find finitely many pairwise disjoint intervals J ,…, J   such that µ*[X\  m |J|] <.
=
1 m j 1 j
Then  m j 1 |J|  µ*[X] –   µ*[X] – 2.
=
j
Write I = [x – , x ] and J = [y, y +]. For each n  {1,…,k}, let D = {j  {1,..,m} : J  I }. Then
n n n n j j j j n j n
{1,…, m} is the disjoint union of D ’s.
n
Note that å j D (f(y + ) – f(y))  f(xn) – f(x –  ) for each n  {1,…,k} since f is increasing.

n
j
j
j
n
n
Summing over n, we get å m = (f(y + ) – f(y))  å k n 1 (f(x ) – f(x –  )), and hence å m j 1 s <
=
=
j 1
n
j
j
n
n
j
j
å k n 1 r , or s( å m j 1 |J|) < r ( å k n 1 |I |). From the earlier estimates we conclude that s(µ*[X] – 2)
=
=
=
j
n
n
< r(µ*[X] + ). Since  > 0 was arbitrary and r < s, we must have µ*[X] = 0.
The conclusion is, it can be extended to more general class of real functions.
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