Is it true that if $f\in L^p(\mu)$, then $|\int_n^{n+1} f(x)dx| \leq \int_n^{n+1} |f(x)|dx$?

Question

Jensen's inequality: Let $\mu$ be a positive measure on a $\sigma$-algebra in a set $\Omega$, so that $\mu(\Omega)=1$. If $f$ is a real function in $L^1(\mu)$, if $a<f(x)<b$ for all $x\in \Omega$, and if $\varphi$ is convex on $(a,b)$, then $$\varphi \bigg(\int_\Omega fd\mu\bigg)\leq \int_\Omega (\varphi \circ f)d\mu$$

$\varphi(x)=|x|$ is convex. Is it true that if $f\in L^p$, then $|\int_n^{n+1} f(x)dx| \leq \int_n^{n+1} |f(x)|dx$?

Answer 1

In general, we have $\displaystyle\int_{X}f(x)d\mu(x)\leq\int_{X}g(x)d\mu(x)$, so $\displaystyle\int_{n}^{n+1}f(x)d\mu(x)\leq\int_{n}^{n+1}|f(x)|d\mu(x)$, and also that $-\displaystyle\int_{n}^{n+1}|f(x)|d\mu(x)=\int_{n}^{n+1}-|f(x)|d\mu(x)\leq\int_{n}^{n+1}f(x)d\mu(x)$, so $\left|\displaystyle\int_{n}^{n+1}f(x)d\mu(x)\right|\leq\displaystyle\int_{n}^{n+1}|f(x)|d\mu(x)$.

One uses Holder's inequality to prove that $L^{p}(\Omega)\subseteq L^{1}(\Omega)$ for finite measure $\Omega$.