i need to show that for all bounded $E\in\cal B_{\mathbb R}$ (while $\cal B_{\mathbb R}$ is the borel $\sigma$-algebra) and for $\mu$ a measure on $\cal B_{\mathbb R}$ and $\mu (E)<\infty$ and $\mu$ holds $\mu (\lambda E)=|\lambda |\mu (E) $ and $\mu (\lambda +E)=\mu (E) $ for some $\lambda\in\mathbb R$ , then there is $0\le c<\infty$ scuch that $\mu (E)=c\cdot m(E)$ while $m$ is the lebesgue measure.
i tried somthing but i am pretty sure that there is a problem with it.
what i did was to define a function $f_E:[0,\infty)\to[0,\infty]$ by $f_E (x)=x\cdot\mu (E)$
the function is well defind since $\mu (E)\in\mathbb R$ and it is easy to see that it is continuos, and $\lim _{x\to\infty}f_E (x)=\infty$ and $\lim _{x\to 0}f_E (x)=0$ then from the intermediate theorem we get there is $\frac 1 c\in\mathbb [0,\infty)$ such that $f_E (\frac 1 c)=m(E)$ thus $\frac 1 c\mu(E)=m(E)$ so we get $\mu(E)=c\cdot m(E)$ so i am pretty sure something is off as i said. i didnt use any actual measure theory and not any of the conditions exept of the fact that both measures give a finite value on $E$. some help will be great!! thanks
Here is one approach (which use the definition of the Lebesgue measure):
Let $c = \mu (0,1)$.
$\mu(a,b) = \mu(0,b-a) = (b-a) \mu (0,1) = c(b-a) = c m(a,b)$, hence $\mu = c m$ for open intervals.
Suppose $A$ is Borel.
We have $mA = \inf\{ \sum_n m I_n | A \subset \cup_n I_n, I_n \text{ open interval} \}$, hence if $A \subset \cup_n I_n$ with $I_n$ open, we have $\mu A \le \sum_n \mu I_n = c \sum_n m I_n$. Taking the infimum gives $\mu A \le c m A$.
Let $J_n = (-n,n)$. We have $\mu J_n = \mu(A \cap J_n) + \mu (J_n \setminus A) \le cm(A \cap J_n) + cm (J_n \setminus A) = c m J_n$, from which it follows that $\mu(A \cap J_n) = cm(A \cap J_n)$. Since $\mu A = \lim_n \mu(A \cap J_n), m A = \lim_n m(A \cap J_n)$, we have the desired result.
Addendum: A standard (although slightly opaque) result in measure theorem is the Dynkin's $\pi-\lambda$ theorem. A $\pi$-system is a collection of sets that are closed under intersection. The theorem can be used to prove the following: If ${\cal P}$ is a $\pi$-system and $\mu_1,\mu_2$ are measures on $\Omega$ that agree on ${\cal P}$ and there are sets $A_n$ such that $A_n \subset A_{n+1}$, $\cup_n A_n = \Omega$ and $\mu_1(A_n), \mu_2(A_n)$ are finite, then $\mu_1,\mu_2$ agree on $\sigma({\cal P})$. (These can be found in most measure/probability books, for example Durrett's "Probability", Theorems 2.1, 2.2.)
In the above example, you can take ${\cal P}$ to be the class of bounded open intervals (which is a $\pi$-system), $\mu_1 = \mu, \mu_2 = c m$, and $A_n = (-n,n)$. Then the measures agree on the smallest $\sigma$-field generated by ${\cal P}$ which is the Borel $\sigma$-field.
The proof above avoids using these results by using the Lebesgue outer measure.