Proof check: continuous function with compact support is $\lambda^d$ integrable

Question

Let $f:\mathbb{R}^d\to \mathbb{R}$ be a continuous function with compact support. Is $f$ lebesgue integrable with respect to $\lambda^d$.

$f$ is continuous $\Rightarrow f$ is measurable. We need to show $\int f ~\text{d}\lambda^d < \infty.$

The image of a compact set under continuous function is compact (Heine-Borel) and the union of compact sets is compact $$f(\mathbb{R}^d)\subseteq \{0\} \cup f(\text{supp}{f})$$ Continuous function on compact set attains its extrema. Let $M=\max\{|\min f|,|\max f|\}$.

So $$\int f(x)\lambda^d(\text{d}x) \leq M\lambda^d\left(f(\text{supp}f)\right)<\infty.$$

Am I right to assume $f(\text{supp} f)$ has finite measure? I think it works in $\mathbb{R}$ since the image of the support has to be a closed interval but what about $\mathbb{R}^m$

This is not homework, its an exam question.

Answer 1

Actually $\text{supp}(f)$ is compact, and $f(\text{supp}(f))$ is compact and hence bounded and it has finite measure: $f(\text{supp}(f))\subseteq[-M,M]$ for some large $M>0$ and $\lambda^{1}(f(\text{supp}(f)))\leq\lambda^{1}([-M,M])=2M<\infty$.