# UMD Analysis Qualifying Exam/Aug05 Real

## Problem 1

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Let be a bounded measurable function on for which there is a constant such that
Show that . |

## Solution 1

editWe will consider two different regions: {x:|f(x)|>1} and {x:|f(x)|<=1}, then show that the integrals of f over these regions are both finite. Since f is bounded by M, we can use the assumption that m(x:|f(x)|>1) < C/1, and the first integral is this bounded by MC.

For the second integral, consider the set E_{n}={x:|f(x)|<1/n}. By assumption, m(E_{n})<C*n^{1/2}. For n>m, E_{n} clearly contains E_{m}, so we can take the sets S_{n}=E_{n+1}\E_{n}, which is then {x:1/(n+1)<|f(x)|<=1/n}. Using the set containment, we can also produce a bound on m(S_{n}). The measure of S_{n} can be larger than C*(n+1)^{1/2}-C*n^{1/2} (up to C*(n+1)^{1/2}), but this extra mass can only be acquired at the expense of m(E_{n}), and would thus reduce the integral of |f|. Since we are interested in maximizing this integral and proving that this maximum is finite, a bound of C*(n+1)^{1/2}-C*n^{1/2} is justified (ie this is the worst case, with each subsequent E_{n} of maximum possible size as 1/n goes to zero).

This gives the integral an upper bound of the sum as n goes from 1 to infinity of (C*(n+1)^{1/2}-C*n^{1/2})/n (by m(S_{n})*(upper bound of |f| on S_{n})). By comparing consecutive terms, we get the sum of C*n^{1/2}*(1/(n-1) - 1/n), for n = 2 to infinity, with a finite term at n=1. After computing the subtraction, each term becomes C*n^{1/2}/(n*(n-1)), which is of order C/n^{3/2}. Hence the sum converges, and gives us an upper bound for the integral.

Note: the argument for the size of S_{n} is correct, but can definitely be expressed better.