Lebesgue Integration: Finite Measure Not Sufficient

[wayneckm]

Hello all,

Here is my question:

Suppose a measureable space (S,\mathcal{S},\mu) with \mu(S) < \infty and f : S \mapsto [0,\infty), this is not yet sufficient to ensure \int_{S} f d \mu < \infty.

Am I correct?

 

[Office_Shredder]

S=(0,1) and f=1/x

 

[LCKurtz]

Yes. Consider the measure defined on R by

\mu(E) =\int_E \frac 1 {1+x^2}\ dx

for Lebesgue measurable E. Let f(x) = 1/x². Then

\int_R \frac 1 {x^2}\cdot \frac 1 {1+x^2}\ dx \ge \int_{-1}^1 \frac 1 {x^2}<br /> \cdot \frac 1 2\ dx =\infty