Let $\{(X_\nu,\mu_\nu):\nu\in\Lambda\}$ be a family of measurable spaces. Is it true that $\bigoplus_2\{L_2(X_\nu,\mu_\nu):\nu\in\Lambda\}$ isometrically isomorphic to $L_2\left(\bigsqcup\{(X_\nu,\mu_\nu):\nu\in\Lambda\}\right)$. It seems to me that desired isometric isomorphism is $$ i:\bigoplus_2\{L_2(X_\nu,\mu_\nu):\nu\in\Lambda\}\to L_2\left(\bigsqcup\{(X_\nu,\mu_\nu):\nu\in\Lambda\}\right):f\mapsto(x_\nu\mapsto f(\nu)(x_\nu))\qquad x_\nu\in X_\nu,\quad\nu\in\Lambda $$ Here is my proof: $$ \Vert i(f)\Vert^2=\int\limits_{\bigsqcup\{(X_\nu,\mu_\nu):\nu\in\Lambda\}}|i(f)(x)|^2 d\left(\sqcup\{\mu_\nu:\nu\in\Lambda\}\right)(x)= $$ $$ \sum\limits_{\nu\in\Lambda}\quad\int\limits_{X_\nu}|i(f)(x_\nu)|^2d\mu_\nu(x_\nu)= \sum\limits_{\nu\in\Lambda}\quad\int\limits_{X_\nu}|f(\nu)(x_\nu)|^2d\mu_\nu(x_\nu)= $$ $$ \sum\limits_{\nu\in\Lambda}\Vert f(\nu)\Vert^2=\Vert f\Vert^2.$$ Hence $i$ is an isometry. The problem is that I used here summation over sets of indices which can be uncountable. I'm not familiar with this notions and how rigorously define them. If you know where I can read about this type of summation with all defintions and theorems give me a link.

My question: Is this proof correct?

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    $\begingroup$ In this thread there's an explanation of how to define the sum of an arbitrarily indexed family of real numbers. $\endgroup$
    – t.b.
    Jan 3, 2012 at 4:38

1 Answer 1


Yes, your argument is correct, if your $\bigoplus$ means the completion that is implicit in what you write.

The issue about possibly-uncountably-many non-negative reals is not serious: only countably-many summands can be non-zero. (Proof: partition a set of positive reals according to which intervals $(1/n,1/(n+1)]$ they lie in. Each interval can contain only finitely-many summands, or the sum would diverge. Thus, there are only countably-many.)


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