Exchangeability of inner product with the integral

Let $(X,\mathcal{E})$ be a measure space. Let $\mu : \mathcal{E} \to \mathbb{R}^+$ be a positive measure. If $f: X \to \mathbb{R}^m$ is measurable and $z \in \mathbb{R}^m$, is it true that: $\int_B \langle f, z \rangle d\mu = \langle \int_B f d\mu, z \rangle$?

I could prove this for simple functions $f$ but not for general functions.

Any help is much appreciated.

Thanks, Phanindra

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Davide: no, I did not show that. Does it lead to a solution to the general problem? – jpv Aug 30 '11 at 11:43
I suggest to consider two cases: $f$ (coordinatewise) integrable or not. In the latter case you'll have trouble, as terms of the form $\infty - \infty$ may appear. The former case should be easy. – t.b. Aug 30 '11 at 11:43
Theo: I apologize I should have added that $f_k$'s are integrable. – jpv Aug 30 '11 at 11:53
No problem and no need to apologize :) You've got your answer quickly, so everyone is happy, I guess. – t.b. Aug 30 '11 at 11:56

Writing $f=(f_1,\ldots,f_m)$ and $z=(z_1,\ldots,z_m)$, both sides are $\displaystyle\sum\limits_{k=1}^mz_k\int_Bf_k\text{d}\mu$, so, yes, they are equal as soon as everything written makes sense, that is, as soon as every $f_k$ is integrable on $B$.

On the LHS, this follows from the definition of the scalar product $\langle\ ,\ \rangle$ and the linearity of the integral. On the RHS, this follows from the definition of the integral of a vector-valued function and the definition of the scalar product $\langle\ ,\ \rangle$.

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Didier: Thanks, I have complicated things in my "proofs". – jpv Aug 30 '11 at 11:49
Didier: My $f_k$'s are integrable, so this answers my question. Thanks again! – jpv Aug 30 '11 at 11:51

Have you any mind for the case that $f$ and $z$ depend on both $‎B$ and $X$ (I mean to the elements of these spaces)?

If not can we put conditions on $z$? for example $z$ becomes sufficiently small.

As I deal with this problem, the answer of my first question is NO, for the second question, I write: We ‎define ‎‎$‎s := \max_{\xi \in B} \{z (\xi , x)\}$‎, ‎by ‎Fubini's ‎theorem ‎we ‎have‎ ‎$\int_{B}\int_{D} g ‎(‎z - s) ‎‎‎dx ‎d‎\mu(\xi) ‎=‎ ‎\int_{B\times D}‎ g ‎‎(‎z - s) ‎‎‎dx ‎‎d‎\mu(\xi)‎‎$‎ ‎‎ ‎$‎‎‎\leq \|(‎z - s)\| \|g\|‎‎‎‎‎‎$‎‎ ‎$‎‎\leq 2 \|s\|‎ \|g\|‎‎$‎‎ ‎as ‎$‎\|‎‎z‎‎\| ‎\rightarrow ‎0‎‎‎‎$ ‎thus ‎‎$\|s\| ‎\rightarrow 0‎‎‎$‎, ‎now one can write‎ ‎‎$‎‎‎\int_{B \times ‎X‎} g z ‎‎‎dx d‎\mu(\xi)‎‎ =‎‎ \int_{B \times ‎X‎} ‎g‎ ‎s ‎‎‎dx ‎d‎\mu(\xi)‎‎ ‎$‎‎ ‎‎$‎= \int‎_{‎X} ‎s \int‎_{B} ‎g‎ d‎\mu(\xi)‎‎ dx = ‎\langle ‎E[‎g]‎ ,s‎ \rangle,‎‎‎$‎ also ‎noting ‎that‎

‎$$‎\langle ‎E[‎g]‎ ,‎E[z] - s‎ \rangle \leq \int_{‎D}\int_{‎B}‎ (z - s) d‎\mu(\xi)‎‎dx ‎\int_{‎D} \int_{B} g ‎‎‎d‎\mu(\xi)‎‎dx ‎‎ ‎‎$$