If $f$ continuous and it's support is compact, why can we use convergence dominated to prove $$\lim\_{\delta\to 1}\int\_{\mathbb R^d}f(\delta x)dx=\int\_{\mathbb R^d}f(x)dx.$$

**My idea is:**

Since $f$ is continuous, $f(\delta x)\to f(x)$ when $\delta\to 1$. Now let denote $S$ it's support. Suppose that $\delta\leq 2$. Then $$\int\_{\mathbb R^d}|f(\delta x)|dx=\int\_{\delta S}|f(\delta x)|dx\leq\int\_{2S}|f(\delta x)|dx.$$

But is $x\mapsto f(\delta x)$ is well defined on $2S$ ? May be I have a problem with the definition. For example, is $f:]0,\infty ]\to\mathbb R$ defined by $$f(x)\begin{cases} \frac{1}{x}-1&x\leq 1\\ 0& otherwise\end{cases}$$
is a continuous function with compact support ? I would says yes since it's support is $[0,1]$, but in other time, it's not define on $0$, so may be it doesn't work. So is the support has to be included in the domain of the function ?