Vtyjkyuk

Suppose that $E subseteq mathbbR$ is measurable and that the measurable functions $f: E to mathbbR$ satisfies $int_E |f| ^p < infty$. If $omega$ is the distribution function of $f$, prove that $lim_lambda to 0 lambda ^p omega(lambda) = 0$.

I'm completely lost. How do I solve this problem?

We define $omega(lambda) = mu(f(x))$, then $lambda^p omega(lambda) to 0$ when $lambda to +infty$ is well known, since
$$lambda^p omega(lambda) leq int_E |f|^p1_geqlambda$$

It's interesting to see the convergence still holds when $lambda to 0$:

Take $lambda_n downarrow 0$, then we remark that

$$int_E |f|^p geq sum_n |lambda_n+1|^pleft(omega(lambda_n+1) - omega(lambda_n))right) = sum_n (lambda_n^p - lambda_n+1^p)omega(lambda_n)$$

Then for any $epsilon >0$, $exists N$ such that for all $k > N$ $$sum_n=k^infty (lambda_n^p - lambda_n+1^p)omega(lambda_n) < epsilon$$

Then remark that $omega(lambda_n) geq omega(lambda_k), forall n geq k$, so

$$sum_n=k^infty (lambda_n^p - lambda_n+1^p)omega(lambda_k) < epsilon$$

i.e. $$lambda_k^pomega(lambda_k)< epsilon, forall k > N$$

It's esay to complete the proof from here(suppose the conclusion is not true, then exists $lambda_n$ such that blabla, and $lambda_n$ has a decreasing subsequence, then apply what's above to get a contradiction.)

Here is another proof, using the identity $$pint_0^infty lambda^p-1 omega(lambda) , d lambda = int_E |f|^p , dm$$ which one can prove with Fubini-Tonelli. If $fin L^p$, then the integral on the left converges; making the change of variables $lambda = 1/t$, we have $$int_0^infty lambda^p-1 omega(lambda) , d lambda = int_infty^0 frac1t^p-1 omegaleft(frac1t right) frac-1t^2 , dt = int_0^infty frac1t^p+1 omegaleft(frac1t right) , dt$$ Since this integral converges, and the function is nonnegative, we have $lim_c to infty int_c^infty t^-(p+1) omega(t^-1) , dt =0$.

On the other hand, we can bound $$int_c^infty t^-(p+1) omega(t^-1) > int_c^infty t^-(p+1), dt cdot omega(1/c) = frac1pc^-p omega(1/c)$$ Taking $ctoinfty$ gives the desired result by squeezing.