Suppose that $E$ is measurable with $m(E)$ $<$ $\infty$.
ii) Show that $\displaystyle \ \ \int_E 2f\,\,\,$ $=$ $2$$\displaystyle \ \ \int_E f\,\,\,$ if $f$ is bounded and measurable.
I told my TA about it in my self study class and he said do not worry about it. We are not going to go into this topic where we would need to solve these types of problems. $ugh$.
Here is what I know from reading:
If $E$ is an element in $M$ we say $E$ is Lebesgue measurable. $M$ will be called Lebesgue measure.
The proof for ii) is very $vague$. I wanted to see the proof of this because I did not learn it. I am new to this and was wondering how we would prove it to be bounded and measurable. This proof might help later with other proofs (even though my TA said do not look at it. $ugh$). That is why I would like to see this.
For 1), start by writing $f$ as a simple function — that is, write down what it means (decomposition of $f$). Then, observe that $\int_E f=\int_\Omega \mathbf{1}_E f$ (where $\Omega$ is the whole space), and $\mathbf{1}_E f$ is also simple (as $E$ is measurable).
For 2), use the theorem relating any function bounded and measurable to simple fucntions (as a limit), and 1).
Edit: For 2): Since $f$ is bounded and measurable, there exists a sequence $(f_n)_{n\in\mathbb{N}}$ of simple measurable functions uniformly converging to $f$.
Fix any $\varepsilon > 0$. Then there exists $N\in\mathbb{N}$ such that for all $n\geq N$, for all $x\in\Omega$, $|f_n(x) - f(x)|\leq \varepsilon^\prime\stackrel{\rm{}def}{=}\frac{\varepsilon}{m(E)}$. We then have that, $\forall n\geq N$,
$\left| \int_E fd\mu - \int_E f_nd\mu \right| \le \int_E \left|f - f_n\right|d\mu \le \int_E \varepsilon^\prime d\mu = \varepsilon^\prime m(E) =\varepsilon$ ($\dagger$) and similarly $\left| \int_E (2f)d\mu - \int_E (2f_n)d\mu \right| \leq 2\varepsilon$ ($\ddagger$)
i.e $\int_E (2f)d\mu < \infty$, $\int_E f d\mu < \infty$ and, more importantly,
$\int_E (2f)d\mu \stackrel{(\ddagger)}{=} \displaystyle\lim_{n\to\infty} \int_E (2f_n)d\mu \displaystyle\stackrel{(1)}{=} 2\displaystyle\lim_{n\to\infty} \int_E f_n d\mu \stackrel{(\dagger)}{=} 2\int_E fd\mu$