Corollary 2

For a mapping g of T into a topological space G to be $\lambda$-measurable, it is necessary and sufficient that it be $\lambda_\alpha$-measurable for every $\alpha \in A$.

The condition is obviously necessary, since $\lambda_\alpha \leq \lambda$ for every $\alpha$ (Ch. IV, §1, No. 3, Prop. 15). Conversely, suppose that $g$ is $\lambda_\alpha$-measurable for all $\alpha$, denote by $\mathscr{K}$ the set of compact subsets K of T such that $g|K$ is continuous, and let L be a compact set such that $L \cap K$ is $\lambda$-negligible for every $K \in \mathscr{K}$. Since the set $\mathscr{K}$ is $\lambda_\alpha$-dense, L is $\lambda_\alpha$-negligible for every $\alpha$ (Ch. IV, §5, No. 8, Prop. 12), hence is $\lambda$-negligible (Cor. 1). It follows that $\mathscr{K}$ is $\lambda$-dense and that $g$ is $\lambda$-measurable (Ch. IV, §5, No. 10, Prop. 15).