The question is the following:

Suppose $f$ is a real-valued Lebesgue measurable function on a set $E\subset \mathbb{R}$ with finite measure. Given $1 > \varepsilon > 0$. Let $E_\varepsilon = \{x:|f(x)|\geqslant \varepsilon\}$. Suppose $$ \frac{1}{\lambda(E)}\int_E |f(x)|\ d\lambda \geqslant 1 \quad \text{and} \quad \frac{1}{\lambda(E)}\int_E |f(x)|^p\ d\lambda \leqslant 1 $$ for some $1<p<\infty$. Show that $$ (1-\varepsilon)^q \lambda(E) \leqslant \lambda(E_\varepsilon) $$ where $1/p+1/q = 1$.

I tried to compute $$ \int_{E \setminus E_\varepsilon} |f| \ d\lambda = \int_{{x \in E:|f(x)| \leqslant \varepsilon}} |f| \leqslant \int_{{x \in E:|f(x)| \leq \varepsilon}} \varepsilon \ d\lambda = \epsilon \cdot \lambda(E \setminus E_\varepsilon)\leqslant\varepsilon \cdot(\lambda(E )-\lambda(E_\varepsilon)) $$ Therefore, we have \begin{align*} \int_{E_\varepsilon} |f| = \int_E |f| - \int_{E \setminus E_\varepsilon}|f| \geqslant \lambda(E) - \varepsilon \cdot( \lambda(E )-\lambda( E_\varepsilon)) \geqslant (1-\varepsilon) \lambda(E) + \lambda(E_\varepsilon) \end{align*} From Holder's inequality, $$ \int_{E_\varepsilon}|f| \leqslant \left(\int_{E_\varepsilon}|f|^p\right)^{1/p} \cdot \left(\int_{E_\varepsilon}|1|^q\right)^{1/q} = \left(\int_{E_\varepsilon}|f|^p\right)^{1/p}\cdot(\lambda(E_\varepsilon))^{1/q} $$ then we have $$ \left(\int_{E_\varepsilon}|f|^p\right)^{1/p}\cdot(\lambda(E_\varepsilon))^{1/q}\geqslant (1-\varepsilon) \lambda(E) + \lambda(E_\varepsilon) $$ which does not imply anything. I know that I have to apply the Holder's Inequality some how from the relationship between $p$ and $q$, but I can't proceed anywhere else. I have no idea how to use the second inequality as it always gives me something from the other direction. Any help and hint are appreciated!!

Edit: I have already figured it out, and I was actually very close to the solution. Here attached the rest of my approach. \begin{align*} (1-\epsilon) \lambda(E) + \lambda(E_\epsilon)& \leq \left(\int_{E_\epsilon}|f|^p\right)^{1/p} \cdot(\lambda(E_\epsilon))^{1/q}\\ &\leq \left(\int_{E}|f|^p\right)^{1/p} \cdot(\lambda(E_\epsilon))^{1/q}\\ &\leq (\lambda(E))^{1/p} \cdot(\lambda(E_\epsilon))^{1/q} \end{align*} and now it suffices to show that the above equation is equivalent to the conclusion $$ (1-\epsilon)^q\lambda(E)\leq \lambda(E_\epsilon) $$ Divide both side by $(\lambda(E))^{1/p}$ and from the fact that $1-1/p = 1/q$ \begin{align*} (1-\epsilon) [\lambda(E)]^{1/q} + \lambda(E_\epsilon)(\lambda(E))^{-1/p}&\leq (\lambda(E_\epsilon))^{1/q} \end{align*} Raise both sides to the $q$-th power, one has \begin{align*} \lambda(E_\epsilon) &\geq [(1-\epsilon) [\lambda(E)]^{1/q} + \lambda(E_\epsilon)(\lambda(E))^{-1/p}]^q \\ &\geq [(1-\epsilon) [\lambda(E)]^{1/q}]^q \\ &\geq (1-\epsilon)^q \lambda(E) \end{align*}

  • $\begingroup$ The prompt is strange. If $E$ is a finite subset of $\Bbb R$ and $\lambda$ is the Lebesgue measure on $\Bbb R$, then $\lambda(E) = 0$. $\endgroup$
    – kobe
    Apr 15, 2020 at 4:47
  • $\begingroup$ @kobe Sorry, there were some typos. Is it clear now? $\endgroup$
    – Tab1e
    Apr 15, 2020 at 4:53
  • $\begingroup$ @kobe Wait, but you argument does not make sense to me, are you claiming that the Lebesgue measure of a finite subset of $\mathbb{R}$ is zero? $\endgroup$
    – Tab1e
    Apr 15, 2020 at 4:54
  • $\begingroup$ @Table a finite subset of $\Bbb R$ is countable and countable sets have measure zero. $\endgroup$
    – kobe
    Apr 15, 2020 at 4:55
  • $\begingroup$ @kobe I see your point. Let's say $E$ is a subset of $\mathbb{R}$ with finite measure. I think it is what it's supposed to mean. $\endgroup$
    – Tab1e
    Apr 15, 2020 at 5:01

1 Answer 1


There seems to be an issue with the assumptions of this problem. Let $dm_E=\frac{1}{\lambda(E)}\mathbb{1}_E(x)\,d\lambda$, where $0<\lambda(E)<\infty$. Then, by Hölder's inequality $$\frac{1}{|E|}\int_E|f(x)|\,dx =\|f\|_{L_1(m_E)}\leq\|f\|_{L_p(m_E)}\|\,\|\mathbf{1}\|_{L_q(m_E)}=\|f\|_{L_p(m_E)}=\Big(\frac{1}{E}\int_E|f(x)|^p\,dx\Big)^{1/p}$$ It is not possible to have $\|f\|_{L_1(m_E)}\geq1$ and $\|f\|_{L_p(m_E)}\leq1$ at the same time, unless $\|f\|_{L_1(m)}=1=\|f\|_{L_p(m)}$, which is rather restrictive. In such a case, we do have that \begin{align} (1-\varepsilon)^q\lambda(E)\leq \lambda(E_\varepsilon),\qquad0<\varepsilon<1\tag{0}\label{zero}\end{align} See estimates below.

There are some bounds that link the measure of the set $E$ with the size of $f$ for more generic functions. Let $\beta_1:=\|f\|_{L_1(m_E)}$ and $\beta_p=\|f\|_{L_p(m_E)}$. Define $E_{\varepsilon \beta_1}:=\{x\in E: |f(x)|>\varepsilon \beta_1\}$, where $0<\varepsilon<1$. An application of Hölder's inequality yields

\begin{align} \beta_1=\int_E|f|\, m(dx) &=\int_E|f|\mathbb{1}_{\{|f|\leq \varepsilon \beta_1\}}\, m_E(dx)+\int_E|f|\mathbb{1}_{\{|f|> \varepsilon \beta_1\}}\, m_E(dx)\\ &\leq \varepsilon\beta_1+\|f\|_{L_p(m_E)}\|\mathbb{1}_{E_{\varepsilon \beta_1}}\|_{L_q(m_E)}\\ &=\varepsilon\beta_1+\beta_p\Big(\frac{\lambda(E_{\varepsilon \beta_1})}{\lambda(E)}\Big)^{1/q} \end{align} Consequently, \begin{align} \beta^q_1(1-\varepsilon)^q\lambda(E) \leq\beta^q_p\lambda(E_{\varepsilon\beta_1})\tag{1}\label{one} \end{align}

Notice that \eqref{zero} ($\beta_1=\beta_p>0$) follows from \eqref{one}.


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