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Let be $u:\Omega \rightarrow \mathbb{R}^n $ a numerical measurable function. Let be $\tau_y u(x) = u(x-y)$. Then, the support($\tau_y u$)=$y$+support($u$).

I am trying: If $x-y = c$ then $\tau_y u(x) = u(x-y)\iff \tau_y u(c+y) = u(c)$. Then support($\tau_y u$) $=\{c+y \in \Omega; u(c)\neq 0\}$. I think that $\{c+y \in \Omega; u(c)\neq 0\} = y + \{c \in \Omega; u(c)\neq 0\}$, but I don't know making a proof?

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If $A$ is a set, the definition of $y+A$ is $\{y + x : x \in A\}$. –  Nate Eldredge Oct 5 '12 at 12:45
    
then, Is my response correctly correctly? –  juaninf Oct 5 '12 at 13:38

1 Answer 1

There's a minor error: the domain of $\tau_y u$ is not $\Omega$, but $y+\Omega$.

Let $S=\{c \in \Omega; u(c)\neq 0\}$. The equality $$\{c+y : c\in \Omega; u(c)\neq 0\} = y + S$$ a tautology, because (as Nate Eldredge said), $y+S$ is defined as as the set of points of the form $y+s$ where $s$ ranges over $S$.

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