Understanding a step in a proof of $L^p(X)\subseteq L^r(X)\subseteq L^1(X)$

Question

I am trying to understand this proof to the question: $L^p$ and $L^q$ space inclusion.

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Here is the linked answer I am reading:

There is a easy way to show that. Suppose that $p<q$ and X a space measure finite. Take any $f\in L^q$. Then, the q-norm is finite. In this way, $$\int_X|f|^p = \int_{f(x)<1}|f|^p + \int_{f\geq1}|f|^p \leq\int1+\int_{f\geq1}|f|^q\leq\mu(X)+||f||_q^q<\infty$$ A counter example just take $$f(x)=\frac{1}{x}$$ for $x\in(0,\infty)$ and Lebesgue measure. Then $f$ belongs to $L^2$ (integral is 1) but not $L^1$.

Here are my questions:

• Why do the first and second inequality in the following hold?

  $$\int_{f(x)<1}|f|^rd\mu\le\int 1d\mu\le\mu(X)$$

• And on what set does one integrate in the notation $\int 1d\mu$?

Answer 1

• In a general measure space $(X,\mathcal{A},\mu)$, the integral notation $\int f\,d\mu$ or $\int f$ means $\int_X f\,d\mu$, which is not a source of confusion when context is clear.
• For any $\mu$-measurable set $B$, $\int_B f\,d\mu=\int_X f1_B\,d\mu$ by definition, where $1_B(x)=1$ if $x\in B$ and $1_B(x)=0$ if $x\not\in B$.
• "$f(x)<1$" means the set $A:=\{x\in X\mid f(x)<1\}$.

• Observe that $$ |f(x)|^r1_A(x)\le 1^r\cdot 1=1,\quad x\in X, $$ and thus $\int_A|f|^r\,d\mu=\int|f|^r1_{A}\le \int 1$.

• $\int 1\,d\mu=\mu(X)$ by definition of integral of the simple functions:

  (1) $\mu(X)<\infty$ by assumption;

  (2) $1=1\cdot 1_X$ is a simple function.

Answer 2

I suppuse that $\{f(x)<1\}$ should be "$\{|f(x)|<1\}:=\{x \in X;|f(x)|<1\}$", then if you integrate over this set, as $|f(x)|<1$, you have $$\int_{\{|f(x)|<1\}} |f|^{r}d\mu \leq \int_{\{|f(x)|<1\}}1^{r}d\mu=\int_{\{|f(x)|<1\}} 1d\mu =\mu(\{|f(x)|<1\})\leq \mu(X).$$