TITLE: Define $ \mu_1 = \int_a^b xdx,$ and inductively: $\mu_k = \int_a^b x d \mu_{k-1}$. What will be $\mu_k?$
QUESTION [7 upvotes]: Define the following sequence of measures given by $\mu_0 = dx$ (Lebesgue measure), $\mu_1([a,b]) = \int_a^b x \;dx, \mu_k([a,b]) = \int_a^b x\; d\mu_{k-1}, [a,b] \subset [0,1].$ What will be $\mu_{k}?$

This problem was posted by our professor as a challenge, and at the moment we don't have Radon Nikodym theorem at disposition. 
Well, what I tried was to define a sequence of functions $\varphi_n$ which I know how to calcule the integral with respect to $\mu_1$ for each term and which converges monotically to $f = x$. So by the monotone convergence theorem, it would follow that:
$$\int_a^b x d\mu_1 = \lim_n \int_a^b \varphi_n d\mu_1$$
The sequence which I tried is given by:
$$ \varphi_n = \sum_{j=0}^{2^n-1}(a+j\cdot \frac{b-a}{2^n})\chi_{[a+j\cdot \frac{b-a}{2^n},\, a+(j+1)\cdot \frac{b-a}{2^n}]}, $$
which basically is a step function defined on and constant at each $2^n$ intervals by the infimum of the function there. The problem, as you may notice, is to calculate this limit:
$\lim_n \sum_{j=0}^{2^n-1} (a+j\cdot\frac{b-a}{2^n}) \mu_1\bigg ( \bigg [a+j\cdot \frac{b-a}{2^n}, a+(j+1)\cdot \frac{b-a}{2^n} \bigg ] \bigg).$
How to proceed with this? Any good suggestions?
Edit
Solution to the hint given in the answer:
Let $f,g = x$. Then exists an increasing sequence $(\varphi_n)$ of simple functions such that $\lim \varphi_n = g, x \in [a,b] =  A.$ Write for each $ \varphi_n = \sum_k a_{k,n}\chi_{E_{k,n}}$ for each $n$. Since $(\varphi_n)$ is an increasing function, by the monotone convergence theorem we have that 
$$ \int_A g d\nu = \lim_n \int_A \varphi_n d\nu = \lim_n \sum_k  a_{k,n}\nu(E_{k,n}) = \lim_n\sum_k a_{k,n} \int_{E_{k,n}}f d\mu = $$
$$\lim_n \int_A \sum_k a_{k,n} f \cdot \chi_{E_{k,n}} d\mu = \lim_n \int_A \varphi_nf d\mu. $$
Now we note that $(\varphi_n f)$ is also an increasing sequence $(f$ is also positive) whose terms are also measurable and positive and such that $\lim_n \varphi_n f = gf $ (as $f$ is bounded on $A$). Hence, once more by the monotone convergence theorem we have
$$ \int_A g d\nu = \lim_n \int_A  \varphi_n f d\mu = \int_A \lim_n \varphi_n f d\mu = \int_A gf d\mu.$$

REPLY [5 votes]: If $\nu A = \int_A f d \mu$, then show that $\int g d \nu = \int_A gf d \mu$ (indicator functions, simple functions, limits).
It follows that $\mu_n[a,b] = \int_a^b x^n dx$.