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

$$\int_R \frac 1 {x^2}\cdot \frac 1 {1+x^2}\ dx \ge \int_{-1}^1 \frac 1 {x^2} \cdot \frac 1 2\ dx =\infty$$