# bounded functions supported on a set E

per Stein's real analysis(P53), if $f$ is a measurable function bounded by $M$ and supported on a set $E$, then there exists a sequence of simple functions, with each one bounded by $M$ and supported on $E$, and such that simple functions converge to $f$ point wise.

It is easy to conceive up a sequence of simple functions bounded by $M$, but how to make them supported on the same set as $f$?

Thank you very much for your help!

• What kind of function is $f?$ Presumably continuous?! – Igor Rivin Jan 8 '14 at 3:37
• I think f is presumed to only be measurable in the context. – user119758 Jan 8 '14 at 3:50
• Whatever it is, you should state it, since the statement is false without assumptions on $f.$ – Igor Rivin Jan 8 '14 at 3:51

Consider $$f_n(x)=\mathrm{sgn}(f(x))\frac{\left\lfloor |f(x)|2^n\right\rfloor}{2^n}$$
The sequence below is for a non-negative measurable function $f$, but you can extend it to a general measurable function by considering its positive and negative parts.
Consider the sequence of simple functions $$f_n = 2^n\chi_{A_n}+\sum_{j = 0}^{2^{2n}-1}j2^{-n}\chi_{B_j}$$ where $A_n = f^{-1}([2^n, \infty))$ and $B_j = f^{-1}([j2^{-n}, (j+1)2^{-n}))$. Note that $f_n$ has the same support as $f$ and $f_n \to f$ pointwise.
• I think we have the same function represented two ways, except I haven't handled the infinite part... except there is no infinite part since $f$ is bounded by $M$. +1 in any case :-) – robjohn Jan 8 '14 at 3:43
• @user119758: I have seen the statement '$f : X \to Y$ is supported on $E$' to mean that $f|_{X\setminus E} = 0$. That's not the same thing as saying $\operatorname{supp}(f) = E$, but rather $\operatorname{supp}(f) \subseteq E$. Do you know if Stein means what I am interpreting the statement to mean, or does he mean that $\operatorname{supp}(f_n) =E$ for every $n$? – Michael Albanese Jan 8 '14 at 8:39