I want to check if the following assertion is true (and my reasoning thereof). Note, this is not a textbook problem, just asking out of curiosity as I self-study convex geometry.
Claim:
Let $K \subseteq \mathbb{R}^{d}$, for some fixed $d \in \mathbb{N}$, be a convex body. That is, $K$ is (by definition) a compact convex set with a non-empty interior. Then $\Lambda_{d}(K) > 0$, where $\Lambda_{d}(K)$ is the volume of $K$ with respect to the $d$-dimensional Lebesgue measure.
Proof (attempt)
First, since $K$ is compact, it is measurable in $\mathbb{R}^{d}$. Moreover, since $K$ has non-empty interior, then there exists some $\mathbf{x} \in \operatorname{int}(K)$, and $\varepsilon > 0$ such that $B(\mathbf{x}, \varepsilon) \subseteq K$. Here $B(\mathbf{x}, \varepsilon)$ is the open Euclidean ball in $\mathbb{R}^{d}$. It then follows by monotonicity of the Lebesgue measure that
\begin{align} 0 < \Lambda_{d}(B(\mathbf{x}, \varepsilon)) \leq \Lambda_{d}(K) \end{align}
That is, $\Lambda_{d}(K) > 0$, as required $\blacksquare$.
Update: The above proof attempt includes an important clarification about the measurability of $K$ from @R. W. Prado and @copper.hat.
Is this reasoning correct? I have used the fact that $\Lambda_{d}(B(\mathbf{x}, \varepsilon)) > 0$, which is true for all $d \in \mathbb{N}$ by the explicit formula here. I believe this claim and my proof seem correct but there may be subtle measurability issues which I've overlooked.
If the statement is true and the reasoning is incorrect, could you please provide the correct reasoning?
