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
 A: 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 $\langle\ ,\ \rangle$.
A: 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
‎$$‎\langle ‎E[‎g]‎ ,‎E[z] - s‎ \rangle \leq \int_{‎D}\int_{‎B}‎ (z - s) d‎\mu(\xi)‎‎dx ‎\int_{‎D} \int_{B} g ‎‎‎d‎\mu(\xi)‎‎dx ‎‎ ‎‎$$
Am I right? In advance thanks for your contribution.
And one more, norms belongs to proper spaces and are not same.
If you have any question I am eager to here. 
