If f is Lebesgue integrable, then $g(x) = \int_x^{\infty} f(y) d \lambda (y)$ is continuous.

Question

This is a true or false question, $f,g$ both $ \mathbb{R} \rightarrow \mathbb{R}$. I think it is right, but in the proof, I have to show for any positive $\epsilon$, there exists positive $\delta$, such that $|x-x_0| \leq \delta$, then $|g(x) - g(x_0)| = |\int_x^{x_0} f(y) d\lambda (y)| \leq \epsilon$, but my question is since it is Lebesgue integrable, I cannot say f is bounded, so how can I bound this integral then, thank you for any help.

Answer 1

To use your approach let $f_n(x)=f(x)I_{\{x:|f(x)| \leq n\}}$. Let $g_n(x)=\int_x^{\infty} f_n(t)d\lambda (t)$. Can you verify that $g_n \to g$ uniformly (using the fact that $\int_{\mathbb R} |f-f_n| \to 0$)? Since $g_n$ is continous for each $n$ and unifrom limits of continuous functions are continuous it follows that $g$ is continuous.