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I am reading a paper.

They define $$ L_{p,q}(Q) = \{ u \in L_p((0,T); L_q(Y)) : u(t, \cdot) = 0 \text{ on } Y \backslash Y_t \text{ for a.e. $t \in (0,T)$}\}$$ with norm $$\lVert u \rVert_{L_{p,q}(Q)} = \left(\int_0^T \lVert u(x) \rVert^p_{L_q(Y_t)} dx\right)^{1\over p}$$ for $p < \infty.$

They write

Since $q < \infty$, the function $x \mapsto \lVert u(x, \cdot) \rVert_{L_q(Y_t)}$ is measurable by Fubini's theorem. Thus, the space $L_{p,q}(Q)$ is well-defined.

Can someone explain this to me? By well-defined I guess they want to show that that norm inside the integral in the norm of the space $L_{p,q}(Q)$ exists. Is that right? I don't see how Fubini's theorem tells us that that function is measureable. And I guess measurability implies that it can be integrated over $(0,T)$ like in the norm?

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How did they define $L_p((0,T);L_q(Y))$? –  user31373 Jul 5 '12 at 18:16
@LeonidKovalev $u \in L_p((0,T); L_q(Y))$ if $\int_0^T{\lVert u \rVert_{L_q(Y)}^p} < \infty.$ It's just the usual Bochner space. –  soup Jul 5 '12 at 18:36
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1 Answer 1

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The issue they deal with is making sense of the condition "$u(\cdot, t)=0$ on $Y\setminus Y_t$ for a.e. $t$". Given that the elements of the Bochner space are equivalence classes of functions from an uncountable set into a set of equivalence classes of functions, it's reasonable to exercise some caution. One way to make this condition precise is to introduce the characteristic function $\chi(x,t)=1$ if $x\in Y_t$ and $=0$ if $x\notin Y_t$. Then we can say that $u\in L_{p,q}$ iff $u=u\chi$ in the Bochner space, and the norm is the usual Bochner norm. However, we need $\chi$ to be measurable in the product $\sigma$-algebra, and this is where we need to know something about $Y_t$ (and apply Fubini).

You did not tell us anything about $Y_t$.

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Thanks for the answer. $Y_t \subset Y$ is a family of bounded domains in $\mathbb{R}^n$ depending on $t \in (0,T).$ The authors say that the usual Bochner spaces are recovered if $Y_t \equiv Y.$ Also, $Q = \cup_{t \in (0,T)} (\{t\} \times Y_t) \subset \mathbb{R}^{n+1}.$ Does this help? PS: the paper is here (math.ualberta.ca/ijnam/Volume-5-2008/Special-08/08.pdf). –  soup Jul 5 '12 at 20:25
@soup The link helps: it says that $Q$ is assumed to be measurable, which you did not mention. Note that $\chi$ I defined above is just the characteristic function of $Q$. This is where they use Fubini (or rather Tonelli): since $\chi_Q$ is measurable in the product space, it follows that a.e. slice is measurable. ... There is nothing difficult here, just the clarity of writing in the Remark leaves something to be desired. –  user31373 Jul 5 '12 at 20:35
Ah, great. Thanks! –  soup Jul 5 '12 at 21:07
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