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.

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

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