Show limits of function $f(x) := \lambda(B \cap (-x,x])$ and there exist a Borel-set $A$ such that $A \subseteq B$ and $\lambda(A)=a$. 
Consider the measure space $(\mathbb R, \mathcal B(\mathbb R), \lambda)$, where $\lambda$ denote the Lebesgue-measure on $\mathbb R$. Let $B \in \mathcal B(\mathbb R)$ and $f:(0, \infty) \rightarrow [0, \infty)$ be given by $f(x) := \lambda(B \cap (-x,x])$,  $\ x \in (0,\infty)$.

I've shown that $f$ is continuous and increasing on $(0,\infty)$.
However I don't know how to formally find the limits: $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow 0} f(x)$. I should probably use what I've shown ?
I've come to notice that $0 \le \lambda(B \cap (-x,x]) \le 2x$ and $\lambda(B \cap (-x,x]) \le \lambda(B)$, but I don't think I should use the sandwich theorem from Calculus here ?
Also I've tried to show that for every real number $a$ in $[0, \lambda(B)]$ there exist a Borel-set $A$ such that $A \subseteq B$ and $\lambda(A)=a$.
Can someone help me out ?
 A: Since $f$ is increasing we may compute $\lim\limits_{x\to\infty} f(x)$ as $\lim\limits_{n\to\infty} f(n)$, where $n$ only runs through the natural numbers. This allows us to use the properties of any measure, namely that

If $\mu$ is a measure and $(B_n)_{n\geq 1}$ is an increasing sequence of sets, i.e. $B_1\subseteq B_2\subseteq\cdots$  then
  $$
\mu\Big(\bigcup_{n\geq 1} B_n\Big)=\lim_{n\to\infty} \mu(B_n).
$$

and

If $\mu$ is a measure and $(B_n)_{n\geq 1}$ is a decreasing sequence of sets, i.e. $B_1\supseteq B_2\supseteq \cdots$ and $\mu(B_1)<\infty$, then
  $$
\mu\Big(\bigcap_{n\geq 1} B_n\Big)=\lim_{n\to\infty}\mu(B_n).
$$

Note that these properties involve only countably many sets which is why we need to deal with the limit of $f$ through a countable sequence. Use the above with $B_n=B\cap (-n,n]$ and $B_n=B\cap (-1/n,1/n]$ respectively.
For the last part use that $f$ is continuous on $(0,\infty)$ with range $(0,\lambda(B))$ to conclude that for every $a\in (0,\lambda(B))$ there is a set of the form $B\cap (-x,x]$ for some $x\in (0,\infty)$ such that $\lambda(B\cap (-x,x])=a$. Lastly, deal with $a\in \{0,\lambda(B)\}$ separately.
A: Use the continuity of lesbegue measure:
$B_n=B\cap (-n,n] \rightarrow B$ as $n\rightarrow \infty$
and $B_n \subset B_{n+1}$
Now you have $\lim_{n\rightarrow\infty} \lambda (B_n) = \lambda (B)$
for $x\rightarrow 0$ use $B_n = B\cap (-\frac1n,\frac1n]\rightarrow \{smthing of measure 0\}$
A: Hint 
Use the continuity of lesbegue measure and intermediate value theorem
