Using the MCT to evaluate the integral of a series I'm studying for my Measure Theory final and I've come across a question that I can't seem to find an answer for.  
For each $n \in \mathbb{N}$ set $E_n:=[n,2n]$ and let $f:\mathbb{R} \to \mathbb{R}$ be given by $\sum_{n\geq 1} 2^{-n} \chi_{E_n}$.  Show that $f$ is a well-defined nonnegative $\mathcal{M}$-measurable function and use the MCT to evaluate $\int_{\mathbb{R}} f d\lambda$.  
I'm not really sure where to get started on this and I appreciate any help.
Edit:  Problem solved.  
Using the definition of Lebesgue integral, MCT and the fact that 
$$
\frac{d}{dx} \sum_{n=1}^{\infty} x^n =\sum_{n=1}^{\infty} nx^{n-1}= \frac{1}{(1-x)^2}
$$ 
for $|x|<1$ we find that
$$
\int_{\mathbb{R}} f d\lambda = \sum_{n=1}^{\infty} 2^{-n}n=4.
$$
Thank you for any help, please let me know if my summary contains errors.  
 A: I leave it to you to show that $f$ is a "well-defined nonnegative $\mathcal{M}$-measurable function."
Let $f_n := \sum_{k=1}^n 2^{-k} \chi_{[k,2k]}$ be the partial sums of $f$. Then
$$\int f_n \mathop{d\lambda} = \sum_{k=1}^n 2^{-k} k.$$
Since $f_1 \le f_2 \le f_3 \le \cdots$, we may apply the MCT to get
$$\int f \mathop{d\lambda} =\int \lim_{n \to \infty} f_n \mathop{d\lambda}=\lim_{n \to \infty} \int f_n \mathop{d\lambda}= \sum_{k =1}^\infty 2^{-k}k<\infty.$$
If you want a closed form for the series, take the derivative of the geometric series $\sum_k 2^{-k}$ and see what you get.
A: The function is well-defined by simple calculus (here you just need to make sure the series converges at each $x\in\mathbb R$).
Clearly $f$ is nonnegative.  Since $f$ is the supremum of the partial sums $\sum _{n\leq k} 2^{-n} \chi_{E_n}$, $k\geq 1$, which are themselves (simple) measurable functions, $f$ is measurable.
Now apply MCT to the sequence of partial sums (which is nondecreasing).
$$\int_{\mathbb R} f d\lambda=\int_{\mathbb R} \sum _{n\geq 1} 2^{-n} \chi_{E_n}=\int_{\mathbb R} \sup_{k\geq 1} \sum _{n\leq k} 2^{-n} \chi_{E_n}=\sup_{k\geq 1} \int_{\mathbb R} \sum _{n\leq k} 2^{-n} \chi_{E_n}=\sup_{k\geq 1} \sum _{n\leq k} \int_{\mathbb R} 2^{-n} \chi_{E_n}$$
$$=\sum _{n\geq 1} \int_{\mathbb R} 2^{-n} \chi_{E_n}$$
Now do some more calculus.
