# is $f$ integrable (measurable)

I'm trying to prove that the function

$$f: \mathbb{R}\mapsto \mathbb{R}:$$ $$f(x) = \begin{cases} 1/(x-1) & \quad \text{if } x \neq 1\\ 0 & \quad \text{if } x=1 \end{cases}$$ is (not) integrable. I wanted to prove that the function was measurable, but this is where I'm already stuck. I'm sure it is but I can't give a good prove. Someone who can help with this?

If I could prove that I wanted to prove that if $$f$$ is integrable then $$\int|{f}|d\lambda$$ is finite. Now I calculate:

$$\lim{t \to \infty}$$ [$$\int_{-t}^{1}1/(x-1)dx$$+$$\int_{1}^{t}1/(x-1)dx$$]

=$$\lim{t \to \infty}$$[$$[-\ln(1-x)]^{1}_{-t}$$+$$[\ln(x-1)]^{t}_{1}$$]

= $$\infty$$ so $$f$$ is not integrable.

I'm just stuck with the first part of my solution. Proving that $$f$$ is measurable.

• Do you mean $f(x)=\frac{1}{x-1}$ if $x\neq 1$? – zugzug Oct 24 at 17:45
• i'm nog sure where you think i wrote something wrong? – questmath Oct 24 at 17:49
• ah yes i see, i changed it – questmath Oct 24 at 17:50

Typically, to show $$f$$ is measurable you need to show that $$\{x: f(x)< c\}$$ is measurable for all $$c$$. (Equivalently, one can change $$<$$ to $$\leq$$, $$\geq$$, or $$>$$). It should be pretty easy to break down $$c$$ into cases and show that these sets amount to intervals, which are measurable.