Average of a Lebesgue Measurable Function

Question

Say I define an indicator function over a set E and demand that this function have an average of 1/2. It is apparently sufficient to ask that the measure m(E) = 1/2. Why is this? I understood Ave(f) to be integral(f)/m(E) which if I calculated correctly would give a value of 1.

Answer 1

It depends on the total mass of $m$. That is, if $m$ is a finite measure on a space $X$, i.e. $m(X)<\infty$, and $f:X\to\Bbb R$ is a bounded measurable function, then we can define the $m$-average of $f$ as $$ \mathrm{A}f :=\frac1{m(X)}\int_X f\mathrm dm. $$ In case $f = 1_E$ is an indicator function, we clearly have $\mathrm Af = m(E)/m(X)$, so that in general you have to have $m(E) = \frac12 m(X)$ to assure that $\mathrm A1_E = \frac12$.