Let E have measure zero Show that if f is a bounded function

Let E have measure zero. Show that if f is a bounded function on E, then f is measurable and integral_E f = 0.

Solution

By chebychev\'s inequality m{E:f>1/n})<=n integral over E f=0 for all n>=1

we conclude that m({E:f>0})=0

since {E:f>0}=Un=1 to infinity {E:f>1/n} is a countable union of measurable zero sets f=0 almost everywhere on E.