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`\Sigma` -> `\sum`, while this is on the front page

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edited Feb 19 at 18:08

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In standard textbooks, the Lebesgue integral is first defined for f$f$ with values in $[0,\infty]$. Note that the value $\infty$ is allowed, and it is important that it can be taken even on a set of positive measure.

You can see why for example in the one-line proof of the Borel-CantelliBorel–Cantelli lemma:

let $\mu$ be finite, and $A_i$ measurable sets so that $\Sigma \mu(A_i)< \infty$$\sum \mu(A_i)< \infty$. Then a.e. x$x$ belongs to only a finite number of $A_i$.

Proof: $\int \Sigma\ 1_{A_i} d\mu = \Sigma\ \mu(A_i) <\infty$$\int \sum 1_{A_i} d\mu = \sum \mu(A_i) <\infty$. The function $x\mapsto \Sigma\ 1_{A_i}$$x\mapsto \sum 1_{A_i}$ is integrable, hence finite a.e.

If you start with a set of functions $f$ with values in $[-\infty,\infty]$, you will have to assume that either $-\infty$ or $\infty$ is taken only on a set of zero measure (so as to prevent the $\infty -\infty$ problem), and if you want it to work both for $f$ and $-f$, you will have to assume that none of $-\infty$, $\infty$ is taken on a set of positive measure, thus ruling out the kind of argument as in the Borel-CantelliBorel–Cantelli lemma. Which is of course unbearable for an analyst, and even more for a probabilist. But perfectly sound from a functional analysis perspective.

In standard textbooks, the Lebesgue integral is first defined for f with values in $[0,\infty]$. Note that the value $\infty$ is allowed, and it is important that it can be taken even on a set of positive measure.

You can see why for example in the one-line proof of the Borel-Cantelli lemma:

let $\mu$ be finite, and $A_i$ measurable sets so that $\Sigma \mu(A_i)< \infty$. Then a.e. x belongs to only a finite number of $A_i$.

Proof: $\int \Sigma\ 1_{A_i} d\mu = \Sigma\ \mu(A_i) <\infty$. The function $x\mapsto \Sigma\ 1_{A_i}$ is integrable, hence finite a.e.

If you start with a set of functions $f$ with values in $[-\infty,\infty]$, you will have to assume that either $-\infty$ or $\infty$ is taken only on a set of zero measure (so as to prevent the $\infty -\infty$ problem), and if you want it to work both for $f$ and $-f$, you will have to assume that none of $-\infty$, $\infty$ is taken on a set of positive measure, thus ruling out the kind of argument as in the Borel-Cantelli lemma. Which is of course unbearable for an analyst, and even more for a probabilist. But perfectly sound from a functional analysis perspective.

In standard textbooks, the Lebesgue integral is first defined for $f$ with values in $[0,\infty]$. Note that the value $\infty$ is allowed, and it is important that it can be taken even on a set of positive measure.

You can see why for example in the one-line proof of the Borel–Cantelli lemma:

let $\mu$ be finite, and $A_i$ measurable sets so that $\sum \mu(A_i)< \infty$. Then a.e. $x$ belongs to only a finite number of $A_i$.

Proof: $\int \sum 1_{A_i} d\mu = \sum \mu(A_i) <\infty$. The function $x\mapsto \sum 1_{A_i}$ is integrable, hence finite a.e.

If you start with a set of functions $f$ with values in $[-\infty,\infty]$, you will have to assume that either $-\infty$ or $\infty$ is taken only on a set of zero measure (so as to prevent the $\infty -\infty$ problem), and if you want it to work both for $f$ and $-f$, you will have to assume that none of $-\infty$, $\infty$ is taken on a set of positive measure, thus ruling out the kind of argument as in the Borel–Cantelli lemma. Which is of course unbearable for an analyst, and even more for a probabilist. But perfectly sound from a functional analysis perspective.

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created May 18, 2010 at 18:49

coudy

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You can see why for example in the one-line proof of the Borel-Cantelli lemma:

let $\mu$ be finite, and $A_i$ measurable sets so that $\Sigma \mu(A_i)< \infty$. Then a.e. x belongs to only a finite number of $A_i$.

Proof: $\int \Sigma\ 1_{A_i} d\mu = \Sigma\ \mu(A_i) <\infty$. The function $x\mapsto \Sigma\ 1_{A_i}$ is integrable, hence finite a.e.

If you start with a set of functions $f$ with values in $[-\infty,\infty]$, you will have to assume that either $-\infty$ or $\infty$ is taken only on a set of zero measure (so as to prevent the $\infty -\infty$ problem), and if you want it to work both for $f$ and $-f$, you will have to assume that none of $-\infty$, $\infty$ is taken on a set of positive measure, thus ruling out the kind of argument as in the Borel-Cantelli lemma. Which is of course unbearable for an analyst, and even more for a probabilist. But perfectly sound from a functional analysis perspective.