Page 363 - DMTH401_REAL ANALYSIS

Unit 30: Riemann's and Lebesgue

d
d
d
(ii) If z   , there is r   such that z – r  [0, 1] . Then there is y  Y such that y ~ z – r and Notes
so there is r   such that y + r = z – r or z = y + r + r. This shows that  =  r d (Y + r).
d
d
By [102](iii) and [102](viii), we conclude that m * L,d [Y] > 0. Let  = m * L,d [Y] and n   be such
d
d
that n > 2 . Choose distinct elements x ,…,x  X. Then å n m * [Y + x ] = n > 2 again by
=
1 n i 1 L,d i
d
[
translation invariance. On the other hand, Y + X  [0,2] and therefore m * L,d  n i 1 (Y+x )] < 2 .
d
=
i
Note The construction above is due to Vitali, and hence the set Y is called a Vitali set.
30.3 About Functions Behaving Nicely Outside a Small Set
There are a few classical results in Analysis with conclusion of the following form: “… the
function has nice behavior outside a small set”. We will consider some such results here.
We know that a function that is the pointwise limit of a sequence of continuous functions may
not be continuous. For instance, f : [0, 1]   given by f(1) = 1 and f(x) = 0 for x < 1 is the pointwise
limit of (f ), where f : [0, 1]   is f (x) = x .
n
n n n
Definition: Let X, Y be metric spaces and let f: X  Y be a function. Then the oscillation (f, x) of
f at a point x  X is defined as (f, x) = lim diam[f(B(x, ))]. Clearly, f is continuous at x iff 
0+
(f, x) = 0.

Task Let X,Y be metric spaces and let f: X  Y be a function. Then the set {x  X: f is
continuous at x} is a G subset of X. [Hint: The given set is equal to  ¥ U , where U = {x  X:
=
 n 1 n n
(f, x) < 1/n}, and U is open.]
n
Let (X,  ) be a complete metric space, (Y,  ) be an arbitrary metric space, and let (f ) be a
1 2 n
sequence of continuous functions from X to Y, converging pointwise to a function f: X  Y. Then
the set {x  X : f is continuous at x} is a dense G subset of X.

Proof: Let  > 0 and D = {x  X : (f, x) > }. We know that D is a closed set. We claim that D is
  
nowhere dense in X. Let U  X be a nonempty open set. We have to find a nonempty open set V
 U such that D  V = Ø.

For n  , let K = {x  X:  (f (x), fj(x))  /8 for every j  n}. Then K is a closed set and X =  ¥ K .
=
n 2 n n n 1 n
The continuity of the distance function  implies that  (f (x), f(x))  /8 for every x  K . Let
2 2 n n
U  X be a nonempty open set with U  U. Since ( U ,  ) is a complete metric space, there is
1 1 1 1
n  N such that U := int[K  U ]  Ø. Let b  U and V  U be an open set with diam[f (V)]
2 n 1 2 2 n
 /8. For any x, y  V, we have  (f (x), f(y))   (f(x), f (x)) +  (f (x), f (b)) +  (f (b), f (y)) +
2 2 n 2 n n 2 n n
 (f (y), f(y))  /8 + /8 + /8 + /8 = /2. Hence diam[f(V)]  /2 and therefore (f, x)  /2 for
2 n
every x  V. This shows D  V = Ø, proving our claim.

The claim implies that D : =  ¥ n 1 D /n is an F set of first category in X. This completes the proof
=
1

since {x  X : f is continuous at x} = X\D, and X is a complete metric space.
We know that the derivative of a differentiable real function need not be continuous. However,
we can say the following.
Let f :    be differentiable. Then there exists a sequence (g ) of continuous functions from
n
 to  converging to f’ pointwise. Consequently, {x  : f is continuous at x} is a dense G subset

of .

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