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In this thread I asked a question about getting started on a problem.

The question is this:

Let $f$ be a function such that $f(x) = \frac {(-1)^n}n$ for $x\in [n, n+1)$.

1) Show that $lim_{n\to \infty} \int_{[1,n]} f$ exists

2) Is $f$ integrable on $[1, \infty)$

Are 1) and 2) essentially asking the same thing? I've proven 1) by basically noting that $f$ is a step function and thus can be written as a simple function, and was able to show there does exist a limit. So does this show that $f$ is Lebesgue integral on $[1, \infty)$?

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Tadpole
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1 Answers1

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The answer is no. Functions that are integrable (having Lebesgue integral) are absolutely integrable, that is if $\int f$ exists then so does $\int |f|$. In your example $\int |f|$ does not exists since it is essentially the same as the sum of the harmonic series which is divergent.

One might say that part (1) corresponds to the conditionally convergent series $\sum\dfrac{(-1)^n}n$.
On the other hand, part (2) (and more precisely $\int |f| $ ) corresponds to the divergent harmonic series $\sum\dfrac1n$.

Alternatively, if we use that $\int_{E} f = \int_{E} f^{+} - \int_{E} f^-$, both $\int_{E} f^{+}$ and $\int_{E} f^-$ are divergent (where $E=[1,\infty)$, and as usual $f^{+}(x)=\max\{f(x),0\}$ and $f^{-}(x)=\max\{-f(x),0\}$).
Indeed $\int_{E} f^{+}$ corresponds to $\frac12+\frac14+\frac16+\cdots=\infty$.
Indeed $\int_{E} f^{-}$ corresponds to $1+\frac13+\frac15+\cdots=\infty$.

Mirko
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  • This makes sense. However, in the context of this definition: $f: E \to \mathbb{R}$ is measurable, Lebesgue integral of $f$ := $\int f = \int_{E} f^{+} - \int_{E} f^-$ provided at least one of the qualities on the right is fine; If both are finite, we say that $f$ is Lebesgue integrable--would I be looking at the fact that one of either quantities on the right is not finite? – Tadpole Apr 16 '15 at 05:07
  • In this case neither $\int f^+$ nor $\int f^-$ is finite. – Mirko Apr 16 '15 at 05:09
  • And this is because the series diverges, which is because $\int f^+$ and $\int f^-$ are positive and therefore we don't have any "negative" sums, which gives us divergent harmonic series? (I hope this makes sense) – Tadpole Apr 16 '15 at 05:11
  • right, I edited my answer along these lines. – Mirko Apr 16 '15 at 05:15
  • Awesome, thanks for the feedback. – Tadpole Apr 16 '15 at 05:16
  • you are welcome! – Mirko Apr 16 '15 at 05:17