# Limit of an integral, as the measure of the region of integration approaches zero

Hi everyone: Let $f$ be a function defined on on open set $D$ of $\mathbb{R}^{N}$, $(n\geq1)$. Suppose that $(\Omega_{\varepsilon})$ is a family of measurable sets in $D$ such that $\lim_{\varepsilon\rightarrow0}\mu(\Omega_{\varepsilon})=0$, where $\mu$ designates the measure of Lebesgue. How would you prove that if $f$ is locally integrable on $D$, then
$$\lim_{\varepsilon\rightarrow0}\int_{\Omega_{\varepsilon}}f(x)d\mu(x)=0?$$ Thanks for your reply.

If $f$ is integrable (not just locally) on $D$, then for any $\epsilon>0$ it can be approximated by a simple function $\psi$ such that

$$\int_{D}|f-\psi|d\mu < \epsilon /2.$$

Let $M$ be the upper bound for $|\psi (x)|$ on $D$.

Then

$$\left|\int_{\Omega_{\epsilon}}fd\mu\right|<\int_{\Omega_{\epsilon}}|\psi|d\mu+\int_{D}|f-\psi|d\mu < \epsilon/2 +M \mu(\Omega_{\epsilon})$$

Locally integrable is not enough. The function $f(x)=1/x$ is locally integrable on $(0,1)$, but its integrals over the sets $\Omega_\epsilon = (\epsilon,2\epsilon)$ do not tend to zero.

If $f$ is globally integrable, then $|f(x)|\,d\mu(x)$ is a finite measure that is absolutely continuous with respect to $\mu$; the assertion follows.

• Very good. Thank you very much. – user127474 Jun 29 '14 at 17:42