I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.

https://www.youtube.com/watch?v=E8FwgGn1e_o

At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"

That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=\cup_{i=1}^{\infty}E_i$ then

$\sum_{i=1}^{\infty}\lvert \nu (E_i) \rvert =\sum_{i=1}^{\infty}\lvert \int_{E}\chi_{E_i}d\nu \rvert=\lvert \int_{E}(\sum_{i=1}^{\infty}\chi_{E_1})d\nu\rvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $\nu$ is a complex measure.

up vote 1 down vote accepted

The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:

Given $E_1,E_2,\ldots$ disjoint with $E=\cup_jE_j$, define $g:X\to\mathbb C$ by $$g=\sum_{j=1}^\infty\overline{\operatorname{sgn}(\nu(E_j))}\chi_{E_j}.$$ Then $|g|\leq1$, and $$\sum_{j=1}^\infty|\nu(E_j)|=\left|\int_Eg\ d\nu\right|\leq\sup\left\{\left|\int f\ d\nu\right|:|f|\leq 1\right\}.$$ Taking the supremum over all such partitions $\{E_j\}$ of $E$, we obtain $\mu_2(E)\leq\mu_3(E)$.

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    I think when you define $g$ is should involve the conjugate of $\operatorname{sgn(\nu(E_j))}$ – Damo Dec 7 at 1:18
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    Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts. – Damo Dec 7 at 1:24
  • You're completely correct, thanks for pointing that out. – Aweygan Dec 7 at 1:42

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