Bounded measurable function $[0, 1] \to \mathbb{R}$ agrees with function of Baire class $2$ outside set of measure zero? Baire class $1$? We say $f_n \to f$ monotonely on $[0, 1]$ if $f_n \to f$ pointwise and $f_1 \le f_2 \le \dots$ or $f_1 \ge f_2 \ge \dots$. A function $f$ is of Baire class $0$ if $f$ is continuous, and of Baire class $n+1$ if it is a monotone limit of functions of Baire class $n$. I have two questions.


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*Does any bounded measurable function $g: [0, 1] \to \mathbb{R}$ agree with a function $f$ of Baire class $2$ outside a set of measure zero?

*Same question, except with  a function of Baire class $1$.


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*Yes. Baire class $1$ includes indicator functions of closed sets, so Baire class $2$ includes indicator functions of $F_\sigma$'s. Since every measurable set agrees with an $F_\sigma$ a.e., the theorem is true when $g = \sigma_E$. A similar argument is holds for step functions, and then for limits of step functions, which give all $g$.

*No. Let $E \subset [1/2, 1]$ be a Cantor set of positive measure, and let$$g(x) = \chi_E(x) - \chi_E\left(x + {1\over2}\right).$$Then $g$ can not agree a.e. with a function $f$ of Baire class $1$, because $f$ is either upper or lower semicontinuous.

