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While reading the motivation of complete measure space on Wikipedia, what I concluded was, completeness is not really necessary when we define on one measure space and it is necessary when we want to measure on product of measure spaces (is it true ?). If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ? I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ? Essentially what is the flaw in the measure without being complete ? Waiting for response. Thanks!

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    One can define product measure spaces without completions. But Lebesgue measure is complete, and if we want two-dimensional Lebesgue measure, which is complete, to be be the product of one-dimensional Lebesgue measure with itself, we have to complete the product.2012-05-20
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    By definition of Lebesgue measure we can see that it is complete , right, because subset of a set of measure zero has measure zero ie. the subset is measurable, right ?? But how would that change if we take a product ?2012-05-20
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    If $N$ has measure zero, then $\lambda^2(N\times\mathbb{R})\lambda(N)\cdot\lambda(\mathbb{R})=0\cdot\infty=0$. So if $B$ is any subset of $\mathbb{R}$, then $\lambda^2(N\times B)$ would be $0$ if $\lambda^2$ were complete. But if $B$ is not measurable, the set $N\times B$ is not in the usual product $\sigma$-algebra. But it is in the completion.2012-05-20
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    @ Michael still left with few doubts, i didn't really understand ur last comment's second line . Can you explain a bit more? :)2012-05-20
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    That a measure space is complete means that every subset of set with measure zero is measurable (and has therefore measure zero too). We know that $\lambda^2(N\times\mathbb{R})=0$ and $N\times B\subseteq N\times\mathbb{R}$ if $B\subseteq\mathbb{R}$. So if the product were complete, $N\times B$ would be in the product $\sigma$-algebra if it were complete. It is not, which is why we complete the product $\sigma$-algebra to get $\lambda^2(N\times B)=0$ for all $B\subseteq\mathbb{R}$.2012-05-20
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    Btw: The Borel $\sigma$-algebra on $\mathbb{R}^2$ does equal the product $\sigma$ of the one-dimensional Borel $\sigma$-algebras! So the motivation Wikipedia gives might not be very convincing.2012-05-20
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    @ Michael one again , why is $N$ x $B$ not in the product $\sigma-algebra$ ?2012-05-20
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    This does actually require some proof. If $N\times B$ were measurable, then all [sections](http://unapologetic.wordpress.com/category/analysis/measure-theory/page/4/) would be measurable and one of these sections is $B$ (the other one is $\emptyset$).2012-05-20

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