How to show $\int_{-1}^{1}\frac{1}{\lambda-f(x)}dx $ is not integrable if $f$ is continuous and $\lambda$ is in the image of $f$

Question

How to show $\int_{-1}^{1}\frac{1}{\lambda-f(x)}dx $ is not integrable if $f$ is continuous and $\lambda$ is in the image of $f$

Let $y$ be s.t $f(y)=\lambda$

By continuity we know that there is a sequence of $\delta_n$ s.t if $|x-y|\leq \delta_n$ then $|\lambda-f(x)|<\frac{1}{n}$ thus $\chi_{[y-\delta_n,y+\delta_n]}n<\chi_{[y-\delta_n,y+\delta_n]}\frac{1}{|\lambda-f(x)|}$

I am not sure how to finish this off as i do not know how fast $\delta_n \to 0$

Answer 1

It's not true. For example, take $f(x) = x^{1/3}$ and $\lambda = 0$.

It is true if $f$ is differentiable.