# Show that $(1-\epsilon)^q \lambda(E) \leq \lambda(E_\epsilon)$

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 $$1. Show that $$(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*}

• The prompt is strange. If $E$ is a finite subset of $\Bbb R$ and $\lambda$ is the Lebesgue measure on $\Bbb R$, then $\lambda(E) = 0$.
– kobe
Apr 15, 2020 at 4:47
• @kobe Sorry, there were some typos. Is it clear now? Apr 15, 2020 at 4:53
• @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? Apr 15, 2020 at 4:54
• @Table a finite subset of $\Bbb R$ is countable and countable sets have measure zero.
– kobe
Apr 15, 2020 at 4:55
• @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. Apr 15, 2020 at 5:01

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}.