Let $u$ a numerical function measurable in $\mathbb {R}^n$. We define the translation of $u$ by the vector $y$ as $(\tau_y u)(x)=u(x-y)$. Show that
$$supp \, (\tau_y u)=y+ supp \, (u).$$
The definition we are using for support is as follows:
Let $\{\mathcal {O}_i\}_{i \in I}$ the collection of all open subsets of $\mathbb {R}^n$ where $u=0$ almost everywhere in $\mathcal {O}_i$. Consider $\mathcal {O}= \bigcup\limits_{i \in I} \mathcal {O}_i$. Then the support of the measurable function $u$ is closed subset of $\mathbb {R}^n$: $$supp \,(u) = \mathbb {R}^n \setminus \mathcal {O}.$$
My attempt to show inclusion:$supp \, (\tau_y u) \subset y+ supp \, (u).$
Let $x \in supp \, (\tau_y u)$. We want to show that $x \in y + supp \, (u)$, i.e, $(x-y) \in supp \, (u)$. Suppose that $(x-y) \notin supp \, (u)$. Then
$$(x-y) \in \mathcal{O}_u=\bigcup\limits_{i \in I} \mathcal {O}_i,$$ where $\{\mathcal {O}_i\}_{i\in I}$ is the collection of all open subsets of $\mathbb {R}^n$ where $u=0$ almost everywhere in $\mathcal {O}_i$. Thus, there is $i_0 \in I$ such that $(x-y) \in \mathcal{O}_{i_0}$, i.e,
$$u(x-y)=\tau_yu(x)=0 \text{ in } \mathcal{O}_{i_0}.$$
Therefore, $x \notin supp \, (\tau_y u)$. Contradiction.
My teacher said that this proof is not correct, due to the fact that $u(x)=0$ does not mean that $x \notin supp \, u$. For example, if $u: \mathbb R \to \mathbb R$, where $u(x)=1-x^2$, if $x \in (-1,1)$ and $u(x)=0$ otherwise, then $u(-1)=0$, but $-1 \in supp \, u=[-1,1]$.
Thank you in advance for your help.