UMD Analysis Qualifying Exam/Aug05 Real

Problem 1

Let $f$ be a bounded measurable function on $R$ for which there is a constant $C$ such that $\forall \epsilon >0,m(x\in R:|f|>\epsilon )<C/{\sqrt {\epsilon }}$ Show that $f\in L^{1}(R)$.

Solution 1

We 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.