Here, I write $\mu$ as the Borel measure on $\mathbb{R^d}$ satisfying $\mu([a_1, b_1] \times \cdots \times [a_d, b_d]) = (b_1 - a_1) \cdots (b_d - a_d)$ ($a_i, b_i \in \mathbb{R}$) and $\lambda$ as the Lebesgue measure on $\mathbb{R}^d$.
My question is: If $f \colon \mathbb{R} \to \mathbb{R}$ is a Borel measurable function, which is obviously Lebesgue measurable, the $\mu$-integrability of $f$ is equivalent to the $\lambda$-integrability of $f$? And more, two integrals $$ \int_{\mathbb{R}} f d\mu, \quad \int_{\mathbb{R}} f d\lambda $$ are the same if $f$ is integrable? This seems true because $\lambda$ is just a completion of $\mu$, but I'm not sure this holds.
More generally, the same assertion like above for a measure space $(X, \mathcal{M}, \mu)$ and the completion $(X, \overline{\mathcal{M}}, \overline{\mu})$ does hold?