Define $f : [0,1] \to \Bbb R$ by $f(x) := 0$ if $x$ is rational, and $f(x) := d^2$ if $x$ is irrational, where $d$ is the first nonzero digit in the decimal expansion of $x$. Show that $\int_{[0,1]} f d m = 95/3$. Here $m$ is the Lebesgue measure
I know that $f$ is simple, but can someone please suggest on how to proceed with this?
On
Consider $S_d$ to be the set of all irrational $x$ numbers of the form $0.0\ldots 0dp_1p_2\ldots$ as their first non-zero digit. Then we can partition $S_d = S_d^1 \cup S_d^2\cup S_d^3\cup ...$ where the upper index indicates at which decimal point the first digit occurs. It is clear that $m(S_d^1) = 0.1$ (it is comprised of the irrationls in the interval $[0.d,0.d+1)$ by much the same argument $m(S_d^2) = 0.01$ and so on, so since they are all disjoint and partition the irrationals in the unit interval $m(S_d) = 0.1111.... = 1/9$.
Finally, since $f$ is constant equal to $d^2$ in $S_d$, the integral becomes:
$$\int_I f\ dm = \sum_{d=1}^9 (\int_{S_d} f\ dm) = \sum_{d=1}^9 {d^2 \over 9} = {95\over 3}$$
(the previous argument to show that all the $S_d$ had the same measure was wrong, sorry for that)
If you want, you can think about this integral geometrically. The function you described is equal almost everywhere to the following step function (or should I say, function of infinite repeating narrower staircases; intervals are not drawn to scale):
As you can see in the picture, what we get for the total area under that step function of infinite repeating steps is:
$\sum \limits_{n = 1}^{\infty} (81 + 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1) \cdot (.1)^{n}$
$= \sum \limits_{n = 1}^{\infty} 285 \cdot (.1)^{n}$
Since $.1 < 1$, we know this series converges, and it converges to:
$\dfrac{\text{first term}}{1 - \text{common ratio }r} = \dfrac{285 \cdot .1}{1 - .1} = \dfrac{28.5}{.9} = \dfrac{95}{3}$, and thus the total area is $\dfrac{95}{3}$.
Note that I used the fact that if two functions are equal almost everywhere, their integrals are equal.