Measurable function is Baire class 2 almost everywhere - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:16:23Z http://mathoverflow.net/feeds/question/62918 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62918/measurable-function-is-baire-class-2-almost-everywhere Measurable function is Baire class 2 almost everywhere Shlomi 2011-04-25T12:43:34Z 2011-04-25T12:50:05Z <p>Let $X$ be a <a href="http://en.wikipedia.org/wiki/Polish_space" rel="nofollow">polish space</a> (separable completely metrizable topological space). Let $m$ be a probability measure on $X$ and $f:X \rightarrow \mathbb{R}$ a measureable function. I want to show that $f$ is equal a.e. to a second Baire class function (limits of limits of bounded continuous functions).</p> <p>I know how to prove this when $f$ is bounded: in that case $f$ is in $L_1(X)$ and there exists a sequence of continuous functions which converge to $f$ a.e. and are uniformly bounded. Then by taking some limits I can deal with the points in which the sequence doesn't converge.</p> <p>My question is, how to prove for any function $f$, not necessarily bounded?</p> <p>Thanks, Shlomi</p> http://mathoverflow.net/questions/62918/measurable-function-is-baire-class-2-almost-everywhere/62920#62920 Answer by Andreas Blass for Measurable function is Baire class 2 almost everywhere Andreas Blass 2011-04-25T12:50:05Z 2011-04-25T12:50:05Z <p>It seems to me that the general case follows from the bounded case. If $f$ is not necessarily bounded, consider $\arctan\circ f$ and represent it as a limit of limits of continuous functions. Then apply $\tan$ to all those functions and get the required representation of $f$. (Since all the limits you care about are pointwise limits a.e., the fact that $\tan$ is only continuous, not uniformly so, does no harm.)</p>