UMD Analysis Qualifying Exam/Aug12 Real

Problem 1Edit

Compute the following limit. Justify your answer.

 

Solution 1Edit

We will use the dominated convergence theorem. First, note that for   and  ,

 

Therefore,

 

and this function is in  , with

 .

Therefore, by the LDCT,

 

Problem 3Edit

Assume   is absolutely continuous on an interval   and there is a continuous function   such that   a.e. Show that   is differentiable at every   and that   everywhere on  .

Solution 3Edit

If   exists, then by definition,  . So we need to show that this limit both exists and is equal to  .

Then by the absolute continuity of  ,  .

Since,   is continuous, then for any   there exists some   such that for  , .

Therefore,

 .

The same argument gives a lower bound, giving us altogether

 . Therefore, the limit exists (i.e.   is differentiable) and the difference quotient goes to  .


Problem 5Edit

Let   be a nonnegative Lebesgue integrable function on  . Denote by   the Lebesgue measure on  .

(i) Prove that, for each  , there exists a   such that

 

(ii) Prove that, for each  , there is a   such that for each measurable subset  :

if  , then  

Solution 5Edit

(i) Fix epsilon greater than zero. Then, consider the sets Sn={x in [0,1] : n-1<=f(x)<n}. The partial sums of the integrals of f over Sn comprise a monotonically increasing sequence of real numbers, bounded by the finite integral of f over [0,1].

Hence, this sequence converges, and the tail of the sequence, which is the integral of f over the set {x in [0,1] : f(x)>= n}, must eventually be less than epsilon for some n.

(ii) Fix epsilon greater than zero. By part (i), there exists some constant c such that, given the set A={x in [0,1] : f(x)>=c}, the integral of f over A is less than epsilon/2. On the complement of A (in [0,1]), f is bounded above by c, and so any set of measure less than epsilon/2c will produce an integral whose value less than epsilon/2.

If m(A) is nonzero, take delta to be the minimum of m(A) and epsilon/2c. Given any measurable E in [0,1] with m(E) less than delta, the portion in A and the portion in Ac will each have integrals with values at most epsilon/2, so the integral of f over E has a value of at most epsilon.

If m(A)=0, then f is bounded almost everywhere on [0,1], and we simply take delta to be epsilon/c.