Corollary 2

For a numerical function $f$ (finite or not) to be measurable, it is necessary and sufficient that $f^+$ and $f^-$ be measurable. The condition is necessary by Cor. 1; it is sufficient because the image A of X in $\overline{\mathbf{R}} \times \overline{\mathbf{R}}$ under the mapping $x \mapsto (f^+(x), f^-(x))$ does not contain the points $(+\infty, +\infty)$ and $(-\infty, -\infty)$, consequently the mapping $(u, v) \mapsto u - v$ is continuous on A.