Limit of integral of Lebesgue integrable functions I'm preparing for an exam in measure theory and found this exercise:

For $p \in [1, \infty]$ find values of $\lambda$ such that $\lim\limits_{\varepsilon \to 0^{+}} \frac{1}{\varepsilon^{\lambda}} \int_{0}^{\varepsilon} f = 0$ for all $f \in L^{p}[0, 1]$.

By Hölder's inequality the limit is 0 if $\lambda$ < $\frac{p-1}{p}$ and $f(x)=x^{-\alpha}$, $\alpha$ < $\frac{1}{p}$ is a counterexample for the case $\lambda$ > $\frac{p-1}{p}$.
But I'm not sure about $\lambda =\frac{p-1}{p}$, can anyone help? 
 A: Let $E_p:=\{\lambda\in\mathbb R: \forall f\in L^p[0,1], \lim_{\varepsilon\to 0^+}\frac 1{\varepsilon^{\lambda}}\int_{[0,\varepsilon]}f(x)dx=0\}$.


*

*First case: $p=1$: solved by martini and thomas.en, we have $E_1=(-\infty,0]$.

*Second case: $1&lt p&lt\infty$:
Indeed, using maps $x\mapsto x^{-\alpha}$ with $\alpha&lt\frac 1p$ we can see that $E_p\subset (-\infty,\frac{p-1}p]$, and Hölder's inequality shows that $(-\infty,\frac{p-1}p)\subset E_p$, so we have to determine whether $\frac{p-1}p\in E_p$. 


Fix a function $f\in L^p[0,1]$, and put $f_n:=f\chi_{\{-n\leq f(x)\leq n\}}\in L^p[0,1]$. We can write 
\begin{align}
\left|\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}f(x)dx\right|&\leq 
\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}\left|f(x)-f_n(x)\right|dx+
\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}\left|f_n(x)\right|dx\\\
&\leq \frac 1{\varepsilon^{\frac{p-1}p}}\lVert f-f_n\rVert_{L^p}\varepsilon^{\frac{p-1}p}+n\varepsilon^{\frac 1p}\\\
&= \lVert f-f_n\rVert_{L^p}+n\varepsilon^{\frac 1p}
\end{align}
so for each integer $n$
$$\limsup_{\varepsilon\to 0^+}\left|\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}f(x)dx\right|\leq \lVert f-f_n\rVert_{L^p}.$$
Since by the monotone convergence theorem we have 
$$\lVert f-f_n\rVert_{L^p}^p=\int_{[0,1]}\chi_{\{|f|\geq n\}}|f|^p\to 0,$$
we conclude that $\frac{p-1}p$ works so for $1&lt p&lt\infty$ we have $E_p=\left(-\infty,\frac{p-1}p\right]$.


*

*Third case: $p=+\infty$. As siminore showed, with $f=1$ we can see that if $\lambda\in E_{\infty}$ then $\lambda&lt1$, and if $\lambda&lt1$ then for $f\in L^{\infty}[0,1]$ we have 
$$\left|\frac 1{\varepsilon^{\lambda}}\int_{[0,\varepsilon]}f(x)dx\right|\leq \lVert f\rVert_{\infty}\varepsilon^{1-\lambda},$$
which converges to $0$ as $\varepsilon\to 0$. So $E_{\infty}=(-\infty,1)$.

