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No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
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RafaNo, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
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Are continuous functions integrable?
yes, every continuous function is integrable.
Is a corner continuous?
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Yes. For every measurable function, f there's a sequence of simple functions Fn that converge to f m-a.e (wich means for each e>0, there's X' such that Fn|x' -->f|x' and m(X\X')<e).
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