# E measurable with m(E) < $\infty$?

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.

• What definition of the integral are you using? – Zev Chonoles May 15 '13 at 3:14
• In a previous question, you said you were self-taught, but now you have a TA ... which is which? It will affect how people respond. – Calvin Lin May 15 '13 at 3:19
• @CalvinLin I am self taught. I just have one TA to this one class I go to. That is it. – 9959 May 15 '13 at 3:32
• Sorry if this caused any confusion. – 9959 May 15 '13 at 3:33
• @9959 The reason is that there are many different (but equivalent) definitions that people can use, which would affect how they approach the problem. Hence, if it's a class, it's best to state what definition they are using. – Calvin Lin May 15 '13 at 3:36

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,

$\int_E (2f)d\mu \stackrel{(\ddagger)}{=} \displaystyle\lim_{n\to\infty} \int_E (2f_n)d\mu \displaystyle\stackrel{(1)}{=} 2\displaystyle\lim_{n\to\infty} \int_E f_n d\mu \stackrel{(\dagger)}{=} 2\int_E fd\mu$

• Thanks I got number 1 while trying to work it out. We did the samething. – 9959 May 15 '13 at 3:53
• I added more defintion to it. I think it still goes with what we did. – 9959 May 15 '13 at 3:54
• My problem is how would we do number 2). Can you please show me. I know I will understand it if you explain or give a proof to me. Thanks. – 9959 May 15 '13 at 3:55
• @Can you show me 2 please? Still not sure about that. I did not learn that part. – 9959 May 15 '13 at 8:14
• @9959: See edit. – Clement C. May 15 '13 at 10:35