UMD Analysis Qualifying Exam/Jan07 Real

Solution 1
Since $$\int_0^1 f(x)\,dx<\infty$$ then $$f(x)<\infty$$ a.e. on [0,1]. This implies that $$\frac{x^n f(x)}{1+x^n}\to 0$$ a.e. on [0,1] since $$\frac{x^n}{1+x^n}\to 0$$ on (0,1). Then by Lebesgue Dominated Convergence, we have $$\lim_{n\to\infty} \int_0^1 \frac{x^n f(x)}{1+x^n}\,dx =\int_0^1 f(x)\,dx. $$

Now to handle the interval $$(1,\infty)$$:

Case 1: $$\int_1^\infty f(x)<\infty$$
For every $$x\in [1,\infty)$$ we have $$\frac{x^n}{1+x^n}<1$$ and increases monotonically to 1. So by the same argument as above, Lebesgue Dominated convergence gives us $$\lim_{n\to\infty} \int_1^\infty \frac{x^n f(x)}{1+x^n}\,dx =\int_1^\infty f(x)\,dx. $$  and we're done.

Case 1: $$\int_1^\infty f(x)=\infty$$
Notice that for every $$x\in (1,\infty)$$ we have $$\frac{x^n f(x)}{1+x^n} > \frac{f(x)}{2}$$, the right-hand-side must necessarily integrate to infinity. So by monotonicity of the integral we have that $$\int_1^\infty \frac{x^n f(x)}{1+x^n}=\infty$$ for all $$n$$. This gives $$\lim_{n\to\infty} \int_0^\infty \frac{x^n f(x)}{1+x^n}\,dx =\int_0^\infty f(x)\,dx =\infty $$ as desired.

Solution 3
Dividing $$f$$ by $$||f||_p$$, we can assume without loss of generality that $$||f||_p=1$$ (similarly for $$g,h$$ with their appropriate norms). Thus we want to show that $$||fgh||_1\leq 1$$. The proof hinges on Young's Inequality which tells us that

$$\int |fgh|\leq \int \frac{|f|^p}{p} + \frac{|g|^q}{q}+\frac{|h|^r}{r}=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1.$$

Solution 5
We claim that $$f$$ can only take on the values 0 and 1. To see this, suppose the contrary, suppose $$f$$ differs from 0 or 1 on a set $$A$$. We can exclude the case $$m(A)=0$$ because otherwise we can modify $$f$$ on a null set to equal an indicator function without affecting the integral.

Then $$||\Chi_{E_n}-f||_1=\int_{\mathbb{R}\setminus A}|f_n-f|\,dm+\int_{A}|f_n-f|\,dm=0+\int_{A}|f_n-f|\,dm$$

On $$A$$, $$|f_n-f|$$ is a strictly positive function. Then for any $$\epsilon>0$$ sufficiently small there exists some $$A_\epsilon\subseteq A$$ with $$m(A_\epsilon)>m(A)-\epsilon$$ such that $$|f_n-f|\geq C_\epsilon$$ on $$A_\epsilon$$ for some positive constant $$C_\epsilon$$. Then $$||\Chi_{E_n}-f||_1\geq C_\epsilon (m(A)-\epsilon)$$.

Thus we have shown that we can obtain a positive lower bound for $$||\Chi_{E_n}-f||_1$$ completely independent of the choice of $$n$$. This contradicts $$\lim_{n\to\infty} ||\Chi_{E_n}-f||_1=0$$. Hence $$f$$ can only assume values 0 and 1 almost everywhere. Since $$f\in L^1(\mathbb{R})$$, then it is certainly measurable. Hence $$E:=f^{-1}(1)$$ is measurable. And we're done.