Let $f \in L^1_{loc}(\mathbb R)$ then show that there exists $\epsilon>0$ such that $\int_{-\epsilon}^{\epsilon}|f(x)|dx<\frac{1}{2}.$

Question

Let $f \in L^1_{loc}(\mathbb R)$ then show that there exists $\epsilon>0$ such that $\int_{-\epsilon}^{\epsilon}|f(x)|dx<\frac{1}{2}.$

I think we need to find $\epsilon>0$ small enough so that the above can happen.

Any help is appreciated. Thank you.

Answer 1

By assumption, $f$ is integrable on $[-1,1]$. Then show that $\int\limits_{-\epsilon}^\epsilon |f|dx\to 0$ for $\epsilon\to0$.

Answer 2

After some explanation of daw's answer:

Since $f \in L^1_{loc}(\mathbb R)$ we can say that $f \in L^1([-1,1]).$ Let us conder an integral $\int_{-1/n}^{1/n}|f(t)|dt=\int_{-1}^{1}\chi_{[{-1\over n},{1\over n}]}(t)|f(t)|dt$.

Let $g_n(t)=\chi_{[{-1\over n},{1\over n}]}(t)|f(t)|,t \in [-1,1].$

Clearly, $|g_n(t)|\leq |f(t)|\ \ \ \forall t \in [-1,1].$ Again $\displaystyle\lim_{n \to \infty} g_n(t)=0$ for all $t \in [-1,1].$ Again all $g_n$ are measurable functions so we can use Dominated Convergence Theorem.

By DCT $\displaystyle \lim_{n \to \infty}\int_{-1}^{1}g_n(t)dt=\int_{-1}^{1}\lim_{n \to \infty }g_n(t)dt.$

Just putting the values of $g_n(t)$ above we can say that $\displaystyle \lim_{n \to \infty} \int_{-1/n}^{1/n}|f_n(t)|dt=0.$

From here we can easily get our result.