If $f\in L^{1}((0,1))$, does it hold that $\lim\limits_{n\to+\infty}\int_{0}^{\frac{1}{n}}f(x)\mathrm{d}x=0$?

Question

I have met this problem while studying some functional and real analysis. I can't use Holder's inequality, since it seems useless. Any idea?

Answer 1

Yes, by the dominated convergence theorem, since \begin{equation*} \int_0^{\frac{1}{n}} f(x) dx = \int_0^1 f(x) \chi_{[0, \frac{1}{n}]} (x) dx, \end{equation*} where $\chi_{[0, \frac{1}{n}]}$ is the characteristic function of the interval $[0, \frac{1}{n}]$. $f \chi_{[0, \frac{1}{n}]}$ converges to zero pointwise as $n \to \infty$, and the integrand is dominated by $|f(x)|$.

Answer 2

You can also use the condition that $f\in L^1(0,1)$ iff for every $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all measurable sets $E\subset (0,1)$, $m(E) < \delta$ implies $|\int_E f(x)\, dx|< \epsilon$. For the problem, fix $\epsilon > 0$ and choose $N > \frac{1}{\delta}$. If $n \ge N$, then $m((0, \frac{1}{n})) = \frac{1}{n} < \delta$, so $|\int_0^{1/n} f(x)\, dx| < \epsilon$. Since $\epsilon$ was arbitrary, $\lim_{n\to \infty} \int_0^{1/n} f(x)\, dx = 0$.