# Proof verification: $\int_{\Omega} \left(cf\right) d\mu = c\int_{\Omega} f d\mu$

This is a proof verification post. I'm attaching the problem and my solution to it. I believe that my solution is too long. Please check it it is correct and also let me know if there are methods by which I can shorten the proof a bit. Thank you.

The Problem : Let $\left(\Omega,\mathcal{F},\mu\right)$ be a measure space and $f,g:\Omega \to \mathbb{R}$ be Borel functions, i.e. $f^{-1}\left(\mathcal{B}\right) \subset \mathcal{F},~g^{-1}\left(\mathcal{B}\right) \subset \mathcal{F},$ where $\mathcal{B}$ denotes the Borel-$\sigma$-field of $\mathbb{R}$.

Show that, if $\int_{\Omega} f d\mu$ exists and $c \in \mathbb{R},$ then $\int_{\Omega} \left(cf\right) d\mu$ exists and $$\int_{\Omega} \left(cf\right) d\mu=c\int_{\Omega} f d\mu$$

My Solution : To show existence, it is enough to show that $cf$ is Borel. We have to show that $\left(cf\right)^{-1}\left(\mathcal{B}\right) \subset \mathcal{F}$. Let us choose (arbitrarily) $B \in \mathcal{B}$. We want to show that $\left(cf\right)^{-1}\left(B\right) \in \mathcal{F}$.

Note that, $\left(cf\right)^{-1}\left(B\right)=\left\{\omega \in \Omega : \left(cf\right)\left(\omega\right) \in B\right\}=\left\{\omega \in \Omega : cf\left(\omega\right) \in B\right\}$

Suppose $c=0$. Either we have $0 \in B,$ or $0 \notin B$. Thus, $\left\{\omega \in \Omega : cf\left(\omega\right) \in B\right\}=\left\{\omega \in \Omega : 0 \in B\right\}=\phi \text{ or } \Omega,$ and hence, $\left(cf\right)^{-1}\left(B\right) \in \mathcal{F}$.

Now suppose $c \neq 0$. Define, $\frac{B}{c}:=\left\{\frac{b}{c} : b \in B\right\}$. Then, \begin{align*} \left(cf\right)^{-1}\left(B\right)&=\left\{\omega \in \Omega : cf\left(\omega\right) \in B\right\}\\ &=\left\{\omega \in \Omega :\exists b \in B, \text{ such that } cf\left(\omega\right)=b\right\}\\ &=\left\{\omega \in \Omega :\exists b \in B, \text{ such that } f\left(\omega\right)=\frac{b}{c}\right\}\\ &=\left\{\omega \in \Omega : f\left(\omega\right) \in \frac{B}{c}\right\}\\ &=f^{-1}\left(\frac{B}{c}\right) \end{align*} If we can show that $\frac{B}{c} \in \mathcal{B}$, we are done.

Now, $B \in \mathcal{B} \implies \exists \text{ a sequence } \left(A_n\right)_{n=1}^{\infty} \text{ of open intervals in } \mathbb{R} \text{ such that } B=\bigcup_{n=1}^{\infty}A_n$

Claim : $\frac{B}{c}=\bigcup_{n=1}^{\infty}\frac{A_n}{c}$. Now,

\begin{align*} a \in \frac{B}{c} &\implies \exists b \in B, \text{ such that } a=\frac{b}{c}\\ &\implies \exists n \in \mathbb{N} \setminus \left\{0\right\}, \text{ and } b \in A_n, \text{ such that } a=\frac{b}{c}\\ &\implies \exists n \in \mathbb{N} \setminus \left\{0\right\}, \text{ such that } a=\frac{A_n}{c}\\ &\implies a \in \bigcup_{n=1}^{\infty} \frac{A_n}{c} \end{align*}

Thus, $$\frac{B}{c} \subset \bigcup_{n=1}^{\infty} \frac{A_n}{c}$$ To show the other side,

\begin{align*} a \in \bigcup_{n=1}^{\infty} \frac{A_n}{c} &\implies \exists n \in \mathbb{N} \setminus \left\{0\right\}, \text{ such that } a=\frac{A_n}{c}\\ &\implies \exists n \in \mathbb{N} \setminus \left\{0\right\}, \text{ and } b \in A_n, \text{ such that } a=\frac{b}{c}\\ &\implies \exists b \in B, \text{ such that } a=\frac{b}{c}\\ &\implies a \in \frac{B}{c} \end{align*} Thus, $$\bigcup_{n=1}^{\infty} \frac{A_n}{c} \subset \frac{B}{c}$$ Hence we have, $$\frac{B}{c}=\bigcup_{n=1}^{\infty} \frac{A_n}{c}$$

The crucial (although trivial) thing is that for all $n \in \mathbb{N} \setminus \left\{0\right\}, A_n \text{ is an open interval in } \mathbb{R} \implies \frac{A_n}{c} \text{ is an open interval in } \mathbb{R}$. And hence, $\frac{B}{c} \in \mathcal{B}$. This completes the existence part.

Now to show that, $$\int_{\Omega} \left(cf\right) d\mu=c\int_{\Omega} f d\mu \tag{1}$$

If $c=0,$ $(1)$ holds trivially.

$$\text{By definition, }L.H.S.=\int_{\Omega} \left(cf\right) d\mu=\int_{\Omega} \left(cf\right)^+ d\mu-\int_{\Omega} \left(cf\right)^- d\mu$$ $$\int_{\Omega} \left(cf\right)^+ d\mu=\sup\left\{\int_{\Omega}\psi d\mu : \psi \text{ is a simple function}, 0 \leq \psi \leq \left(cf\right)^+\right\}$$ Suppose, $c>0$. Then, $\left(cf\right)^+ = \max\left\{cf,0\right\} = c \max\left\{f,0\right\} = cf^+$

\begin{align*} \int_{\Omega} \left(cf\right)^+ d\mu&=\sup\left\{\int_{\Omega}\psi d\mu : \psi \text{ is a simple function}, 0 \leq \psi \leq cf^+\right\}\\ &=\sup\left\{\int_{\Omega}\psi d\mu : \psi \text{ is a simple function}, 0 \leq \frac{\psi}{c} \leq f^+\right\}\\ &=\sup\left\{\int_{\Omega}c\cdot\frac{\psi}{c} d\mu : \frac{\psi}{c} \text{ is a simple function}, 0 \leq \frac{\psi}{c} \leq f^+\right\}\\ &\left[\psi \text{ is a simple function } \implies \frac{\psi}{c} \text{ is a simple function } \right]\\ &=\sup\left\{\int_{\Omega}c\psi^* d\mu : \psi^* \text{ is a simple function}, 0 \leq \psi^* \leq f^+\right\}\\ &=\sup\left\{c\int_{\Omega}\psi^* d\mu : \psi^* \text{ is a simple function}, 0 \leq \psi^* \leq f^+\right\}\\ &=c\sup\left\{\int_{\Omega}\psi^* d\mu : \psi^* \text{ is a simple function}, 0 \leq \psi^* \leq f^+\right\}\\ &=c\int_{\Omega} f^+ d\mu \end{align*}

In the same way, $\int_{\Omega} \left(cf\right)^- d\mu = c\int_{\Omega} f^- d\mu,$ and hence,

$$\int_{\Omega} \left(cf\right) d\mu=\int_{\Omega} \left(cf\right)^+ d\mu-\int_{\Omega} \left(cf\right)^- d\mu=c\int_{\Omega} f^+ d\mu-c\int_{\Omega} f^- d\mu=c\int_{\Omega} f d\mu$$

Similarly, if $c<0,$ we have $\left(cf\right)^+ = \max\left\{cf,0\right\} = c\min\left\{f,0\right\} = -c\max\left\{-f,0\right\} = -cf^-,$ and hence, $\int_{\Omega} \left(cf\right)^+ d\mu = -c\int_{\Omega} f^- d\mu$.

Also in this case, $\left(cf\right)^- = \max\left\{-cf,0\right\} = -c\max\left\{f,0\right\} = -cf^+,$ and hence, $\int_{\Omega} \left(cf\right)^- d\mu = -c\int_{\Omega} f^+ d\mu$.

Thus we have $$\int_{\Omega} \left(cf\right) d\mu=\int_{\Omega} \left(cf\right)^+ d\mu-\int_{\Omega} \left(cf\right)^- d\mu=-c\int_{\Omega} f^- d\mu+c\int_{\Omega} f^+ d\mu=c\int_{\Omega} f d\mu$$

Hence, in all the cases, $(1)$ holds. Hence the proof.

Seems like you are doing a lot of extra work. How about recalling that the integral of a simple function $\psi(x) = \sum \limits_{i=1}^n a_i \chi_{A_i} \left( x \right)$ is $\sum \limits_{i=1}^n a_i \mu(A_i)$. In this case the proof is trivial. Then you can wrap up by e.g. recalling that any measurable $f$ is a pointwise limit of a monotone sequence of simple functions and then applying monotone convergence theorem.