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Simon Henry
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Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ fromby a Lebesgue negligible set.

$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.

Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.

Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ from a Lebesgue negligible set.

$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.

Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.

Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ by a Lebesgue negligible set.

$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.

Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.

Source Link
Simon Henry
  • 42.4k
  • 5
  • 107
  • 205

Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ from a Lebesgue negligible set.

$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.

Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.