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Exercsie 2.12.37 from Measure Theory Volume 1 (Bogachev)

Let $J_n$ be a sequence of disjoint intervals in $[0,1]$, convergent to the origin, $\left|J_n\right|=4^{-n}$, and let $f=n^{-1} /\left|J_{2 n}\right|$ on $J_{2 n}, f=-n^{-1} /\left|J_{2 n+1}\right|$ on $J_{2 n+1}$, and let $f$ be zero at all other points. Show that $f$ is Riemann integrable in the improper sense, but is not Lebesgue integrable.

I know how to show that it is Riemann integrable. Take an interval of the form $[0,\epsilon]$ for which for any partition $P$ of $[0,1]$ we will have (restricted to this interval), $U(P) - L(P) < 2\epsilon$ (since $f$ is between $-1$ and $1$). For the interval $[\epsilon, 1]$ we can use the fact that a function with finitely many discontinuities is riemann integrable (e.g. How do I prove that a function with a finite number of discontinuities is Riemann integrable over some interval?)

However, I'm not sure how to prove this isn't Lebesgue integrable. My belief is that this function is Lebesgue integrable since we can define $g_N$ to be equal to $f$ on $J_1 \cup \cdots \cup J_N$ and $0$ everywhere else. This is a step function with finitely many discontinuities and has lebesgue integral equal to $$-1\cdot 4 + \frac12 \cdot 4^2 - \frac13\cdot 4^3 + \cdots + \frac{(-1)^N}{N}\cdot 4^N$$ which goes to infinity?

Some guidance would be greatly appreciated!

Mittens
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Math_Day
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1 Answers1

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In order to be Lebesgue measurable you need both $\int f^+ \mathrm d\mu$ and $\int f^- \mathrm d\mu$ to be finite. In your case $$\int f^+ \mathrm d\mu=\int_0^1 \sum_{n=1}^\infty\frac1{n|J_{2n}|}\mathbf1_{J_{2n}}(x)\mathrm dx=\sum_{n=1}^\infty\frac1{n|J_{2n}|}\int_0^1\mathbf1_{J_{2n}}(x)\mathrm dx=\sum_{n=1}^\infty\frac1n=\infty$$ And similarly for $f^-$. According to Riemann instead, you could say that the integral is well defined by considering $$\lim_{y\downarrow0}\int_y^1 f(x)\mathrm dx.$$

Andrea Aveni
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