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If $f$ is a bounded function (not necessarily measurable), is it true that we can find a sequence of simple functions $\{\varphi_n\}$ such that $\varphi_n\rightarrow f$?

I wonder this because in the definition of Lebesgue integral, we define $\int f$ to be the limit of $\int \varphi_n$ when $\varphi_n\rightarrow f$, and we only assumed $f$ to be bounded and supported on a set of finite measure, but we didn't assume that it was measurable, then can we always find {$\varphi_n$} to define the Lebesgue integral of $f$?

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  • $\begingroup$ I wonder if the second sentence could be broken into 2 sentences? Or, is there a missing word, perhaps "if" between "but" and "we"? $\endgroup$ – Jonas Meyer Jan 1 '13 at 7:14
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The limit of a sequence of measurable functions is measurable. If $f$ is not measurable, then there can be no sequence of measurable functions converging pointwise to it.

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  • $\begingroup$ so the definition of Lebesgue integral must be for the measurable functions right? $\endgroup$ – Alex Dec 31 '12 at 18:11
  • $\begingroup$ Yes. It doesn't make sense otherwise. $\endgroup$ – JSchlather Dec 31 '12 at 18:16
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If $f$ is a bounded function(not necessarily measurable), is it true that we can find a sequence of simple functions $\{\varphi_n\}$ such that $\varphi_n\rightarrow f$?

The question is a little ambiguous, because from context one might guess that your simple functions are assumed to be measurable, but this is not a universal assumption and wasn't made clear.

  • If a simple function is defined to be a function that has finitely many values, then the answer is yes, but this doesn't help you define integrals without the measurability condition.
  • If a simple function is assumed to be measurable, which is equivalent to adding the condition that each set on which the function takes one of its finitely many values is measurable, then the answer is no for the reason Jacob Schlather gave.
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