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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, $n<n+1$, 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?

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3  
What have you tried? Any guesses as to the answers? –  Jonas Meyer Nov 9 '12 at 21:25
    
How would you evaluate an integral with respect to counting measure that does not converge absolutely? –  Michael Greinecker Nov 10 '12 at 4:06
    
my attempt is now included with the question –  Jake Casey Nov 10 '12 at 17:17

2 Answers 2

To see that convergence of $a_n$ does not do much for the integrability of $f$, you should construct a function $f$ such that $a_n=0$ for all $n$ but $\int_1^\infty f(x)\,dx$ diverges.

Suggestion: divide each interval $[n,n+1)$ in two halves, make $f$ large and positive one on half, and equally large with opposite sign on the other.

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Consider $f(x) = \sin(2 \pi x)$.

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