# Lebesgue Integral

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• Nov 8th 2009, 11:38 PM
problem
Lebesgue Integral
Let the sequence of integrable functions $\displaystyle ( f_n )$ be defined on the interval $\displaystyle [0,\infty)$ in which $\displaystyle f_n = \dfrac{1}{n} 1_{[0,n]}$ on $\displaystyle [0,\infty)$.
Prove that there is no integrable function that dominates the sequence.

My work:
Suppose there exist a function$\displaystyle f$that dominates the sequence.
Hence,
$\displaystyle \int_{[0,\infty)} f dm$
$\displaystyle >\int_{[0,n]} f dm$
$\displaystyle >\sum_{k=1}^{n} \int_{[k-1,k)} f dm$
$\displaystyle >\sum_{k=1}^{n} \int_{[k-1,k)} \dfrac{1}{n} dm$

I was stuck here.
Can anyone help me to proceed?
Or if this is not the way to prove,can anyone please show me the correct way?
Thanks.
• Nov 9th 2009, 01:47 AM
Enrique2
If there existed such a dominating function, you could apply Lebesgue Dominated Convergence theorem, but

$\displaystyle \lim_nf_n=0$ pointwise (even uniformly!)

and

$\displaystyle \int f_n=1$ for each $\displaystyle n$

Hence we can't interchange limit and integral, a contradiction with having the condictions to apply Lebesgue theorem.