Measure Theory and $L^{p}$ spaces

Question

I have the two following very simple questions regarding measure theory that I want to show:

1. If $f \in L^{p}(X, \mathcal{M}, \mu)$ for $1 \leq p < \infty$, then $f < \infty$ $\mu$-almost everywhere (or, $f < \infty$ for all $x \in X$ except for a $\mu$-null set $E \in \mathcal{M}$).

2. If $f=0$ $\mu$-almost everywhere (for a measure space $(X, \mathcal{M}, \mu)$), then $\int f \mathrm{d\mu}=0$

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1. For this, I'm not so sure on how to begin. If $p=\infty$, the result is trivial by the definition of the essential supremum. However, for the other case I don't know how to approach it.

2. The converse is straightforward and it follows by Tchebyshev's inequality, but I don't know where to begin in order to prove the converse statement.

Thanks for the help.

Answer 1

1. Let $E_n=\{x:|f|>n, \:n\in \Bbb{N}\}$. Then by Chebyshev's inequality $$ \mu(E_n)\leqslant \frac1{n^p}\int_{X}|f|^pd\mu\quad\text{and }\quad\lim_{n\to\infty}\mu(E_n)=0 $$ Let $E=\bigcap_{n=1}^{\infty}E_n$. Then $E=\{x:|f|=\infty\}$. Since $E_1\supset\cdots\supset E_n\supset\cdots\supset E$, by Monotone class theorem $$ \mu(E)=\lim_{n\to\infty}\mu(E_n)=0 $$ So $\mu(\{x:|f|<\infty\})=\mu(E^c)$, i.e. $|f|<\infty$ a.e.

2. Let $A=\{x:|f|\ne 0\}$. Then $\mu(A)=0$. For any $C\subset X$, $\mu(C\cap A)=0$. So $$ \int_{C}|f|^pd\mu=\int_{C\cap A}|f|^pd\mu+\int_{C\cap A^c}|f|^pd\mu=0 $$

Answer 2

1. If $f=\infty$ on a set of positive measure, say $E$, then

$$\int |f|^p \ge \int |f|^p 1_E =\infty.$$

2. $|\int f|\le \int|f| =0$ because $|f|=0$ $\mu$-a.e. and for $g\in L^+$ this can be shown using the definition of the integral (i.e., $\sup\{\int \phi : \phi \text{ is simple s.t. }0\le\phi\le g\}$).

   Also the converse is not true in general (only for nonnegative functions).