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For a measurable function, $f$, on $[1, \infty)$ which is bounded on bounded sets, define $a_n = \int_n^{n+1} f$ for each natural number $n$. Is it true that $f$ is integrable over $[1, \infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges? Is it true that $f$ is integrable over $[1, \infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges absolutely?

Here is what I am thinking:

Let $\sum_1^\infty a_n$ be convergent. $\sum_1^\infty a_n$ = $\sum_1^\infty (\int_n^{n+1}f)$. Since n are natural numbers, $n, then $\sum_1^\infty (\int_n^{n+1}f)$ = $\int_1^2 f + \int_2^3 f + $... = $ \int_1^\infty f$. Can we say then, that since f is measurable on $[1, \infty)$ and bounded that ${f_n} -> {f}$ a.e. and by the Lebesgue Dominated Convergence Theorem, f is integrable?

Conversely, let f be integrable over $[1, \infty)$.

Given the above is true. To show for absolute convergence we would need to show for $\sum_1^\infty |a_n|$.

Am I headed in the right direction?

  • 0
    How would you evaluate an integral with respect to counting measure that does not converge absolutely?2012-11-10
  • 0
    my attempt is now included with the question2012-11-10

2 Answers 2