$L^p$ space inequality

Question

Question Let $ (X, \mathcal{E}, \mu) $ be a measure space with $\mu(X)=1$. Assume that $1<p<\infty$ and let $1<q<\infty$ be such that $\frac{1}{p} + \frac{1}{q}=1$. Show that if $f\in \mathcal{L}^p$ with $\lVert f \rVert_p>0$ and $0<\lambda<\lVert f \rVert_1 $, then show that $$\mu\left( \left\{x\in X \bigg\vert \lvert f(x)\rvert\geq \lambda \right\} \right) \geq \left(\frac{\lVert f \rVert_1 - \lambda}{\lVert f \rVert_p} \right)^q .$$ Hint Note that if $M = \mu\left( \left\{x\in X \bigg\vert \lvert f(x)\rvert\geq \lambda \right\} \right) $, then $$\int \lvert f\rvert d\mu = \int \lvert f \rvert 1_M d\mu + \int \lvert f \rvert 1_{X \backslash M} d\mu. $$ Attempt I am trying to solve this question. By using hint, I am able to obtain Chebyshev's inequality which gives a upper bound for $M = \mu\left( \left\{x\in X \bigg\vert \lvert f(x)\rvert\geq \lambda \right\} \right) $. To show this inequality, how should I start? Thank you!

Answer 1

Use the definition of $M$ in the integral with $X\setminus M$ and Hölder in the other integral, so:

$$ \lVert f\rVert_{1} \leq \lVert f \rVert_{p} \lVert 1_{M}\rVert_{q} + \int \lambda 1_{X \setminus M} d\mu \leq \lVert f \rVert_{p} \left(\mu\left(M\right)\right)^{\frac{1}{q}} + \lambda\left(1-\mu\left(M\right)\right), $$

which give us

$$ \lVert f\rVert_{1} - \lambda \leq \lVert f \rVert_{p} \left(\mu\left(M\right)\right)^{\frac{1}{q}} - \lambda \mu\left(M\right) \leq \lVert f \rVert_{p} \left(\mu\left(M\right)\right)^{\frac{1}{q}} $$

and from this you can conclude.

Answer 2

$\int \lvert f\rvert d\mu = \int \lvert f \rvert 1_M d\mu + \int \lvert f \rvert 1_{X \backslash M} d\mu \le \underbrace{\lVert f \lVert _{p}\ \mu(f \ge \lambda)^{\frac{1}{q}}}_{\text{Hölder's inequality}} + \lambda$

Rearranging gives the result.