Prove: If $f_1$ and $f_2$ are integrable, then $f_1\vee f_2$ is integrable over each $A\in\mathscr{A}$.

Question

I need to prove the following result:

Let $(X,\mathscr{A},\mu)$ be a measure space. If $f_1$ and $f_2$ are integrable, then $f_1\vee f_2$ is integrable over each $A\in\mathscr{A}$.

Here is my attempt:

Proof$\quad$ Since $f_1$ and $f_2$ are integrable, by Exercise 2.3.1 we have $f_1\vee f_2$ is integrable. Thus Proposition 2.3.8 implies $|f_1\vee f_2|$ is integrable. Since $|(f_1\vee f_2)\chi_A|\leq|f_1\vee f_2|$, Proposition 2.3.4 implies $$ \int|(f_1\vee f_2)|d\mu \leq \int|f_1\vee f_2|d\mu <+\infty. $$ Thus $|(f_1\vee f_2)\chi_A|$ is integrable and so is $(f_1\vee f_2)\chi_A$ (again by Proposition 2.3.8). So $f_1\vee f_2$ is integrable over any $A\in\mathscr{A}$.

Is this proof correct? Moreover is it necessary to prove it in such way? Thanks a lot for helping me check it out!

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Definition and results in the question and my attempt:

Definition$\quad$ Let $f$ and $g$ be $[-\infty,+\infty]$-valued functions having a common domain $A$. The maximum and minimum of $f$ and $g$, written $f\vee g$ and $f \wedge g$, are the functions from $A$ to $[-\infty,+\infty]$ defined by \begin{align*} (f \vee g)(x) = \max\{f(x),g(x)\} \end{align*} and \begin{align*} (f \wedge g)(x) = \min\{f(x),g(x)\}. \end{align*} Equivalently, we can define $f \vee g$ by \begin{equation*} (f \vee g)(x) = \begin{cases} f(x) &\text{if $f(x)>g(x)$ and,}\\ g(x) &\text{otherwise,} \end{cases} \end{equation*} and $f \wedge g$ by \begin{equation*} (f \wedge g)(x) = \begin{cases} f(x) &\text{if $f(x)<g(x)$ and,}\\ g(x) &\text{otherwise.} \end{cases} \end{equation*}

Exercise 2.3.1$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, and let $f$ and $g$ belong to $\mathscr{L}^1(X,\mathscr{A},\mu,\mathbb{R})$. Then $f\vee g$ and $f\wedge g$ belong to $\mathscr{L}^1(X,\mathscr{A},\mu,\mathbb{R})$.

Proposition 2.3.8$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, and let $f$ be a $[-\infty,+\infty]$-valued $\mathscr{A}$-measurable function on $X$. Then $f$ is integrable if and only if $|f|$ is integrable. If these functions are integrable, then $|\int fd\mu|\leq\int|f|d\mu$.

Proposition 2.3.4$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, let $f$ and $g$ be $[0,+\infty]$-valued $\mathscr{A}$-measurable functions on $X$m and let $\alpha$ be a nonnegative real number. Then

1. $\int \alpha fd\mu = \alpha\int fd\mu$,
2. $\int(f+g)d\mu = \int fd\mu+\int gd\mu$, and
3. if $f(x)\leq g(x)$ holds at each $x$ in $X$, then $\int fd\mu \leq \int gd\mu$.

Answer 1

Your proof looks correct to me.

Using Proposition 2.3.4 c) in each step, you find $$\int |(f_1\vee f_2)\chi_A|d\mu\le \int| f_1\vee f_2|d\mu\le \int |f_1|+|f_2|d\mu$$

Using Proposition 2.3.4 b), it holds

$$\int |f_1|+|f_2|d\mu=\int |f_1|d\mu+\int |f_2|d\mu$$

Since $f_1$ and $f_2$ are integrable, you find with with Proposition 2.3.8 that

$$\int |f_1|d\mu+\int |f_2|d\mu<\infty$$

This yields

$$\int |(f_1\vee f_2)\chi_A|d\mu<\infty$$