Show the function is integrable and find the integral - somewhat complex question

Question

We are given $Q = [0,1]$x$[0,1]$

We are also given the function $f(x,y) = (\frac{1}{10})^n$ where $\frac{1}{2^{n+1}} < \max(x,y) \leq \frac{1}{2^n}, (n=0,1,2,...)$ and $f(0,0)=0$.

Show that $f$ is integrable over $Q$ and find $\int \int_Q f dxdy$

Firstly, I'm having trouble understanding $f$. but if i understood it. I would try to show that the set of points from $Q$ at which $f$ is not continuous is negligible. Would that show that $f$ is integrable? and even then how would I find the actual value of the integral

Answer 1

Write $Q$ as disjoint union of two sets $A$ and $B$ where $f$ is continuous in $A$, and $f$ is not continuous in $B$. Note that $B$ has measure zero, hence it is enough to integrate over $A$, Write $A$ as disjoint union of sets $A_i$ where $f$ is constant on each $A_i$. Note that on such $A_i$ it is easy to integrate the constant function. Now sum that values and see that $f$ is integrable .

Answer 2

After thinking it through and consulting with a friend I think I managed to solve it.

Thanks to GA316 for hinting me in the right direction.

Would love someone to review this answer.

1) The continuity problem arises when we are on the borderline of shifting $n$. That happens when either $x$ is a power of $\frac{1}{2}$ or $y$ is. Meaning, let $B$ be the set of points in $Q$ at which $f$ is not continuous. Then $B= \{(x,y) \in Q s.t \max(x,y)=\frac{1}{2^k}, k \in \mathbb Z\}$. If we look at this set, we will see its just straight lines, for example $(\frac{1}{2},y)$ where $y \leq \frac{1}{2}$ or $(x,1)$ where $x \leq 1$. So $B$ is just a union of short straight lines, like straws. A straight line is a zero measure set in $\mathbb R^2$ (since it has no volume, it doesnt create space) and so $B$ is in itself zero measure. Meaning $f$ is integrable on $Q$.

2) Let's define $A=Q\setminus B$. The set of points in $Q$ in which $f$ is continuous. Since $B$ is of zero measure, $\int_Q f = \int_A f$.

We will find $\int_A f$.

Since f is a constant in segments, we could find the integral of each segment and since they do not intersect, we could just add them to get the desired result. Meaning, if $A_i \subset \mathbb R^2$ is the segment over which $f=\frac{1}{10^i}$ then:

$\int_A f = \sum_{i=0}^{\infty} \int_{A_i} f$.

Let's find $\int_{A_i} f$:

if $f=\frac{1}{10^i}$ then that means $\frac{1}{2^{i+1}} < \max(x,y)\leq \frac{1}{2^i}$ When does this happen?

a) when $x \in (\frac{1}{2^{i+1}},\frac{1}{2^i}]$ and $y\in (\frac{1}{2^{i+1}},\frac{1}{2^i}]$

b) when $x \in (\frac{1}{2^{i+1}},\frac{1}{2^i}]$ and $y \in [0,\frac{1}{2^{i+1}}]$

c) when $x \in [0,\frac{1}{2^{i+1}}]$ and $y \in (\frac{1}{2^{i+1}},\frac{1}{2^i}]$

for symmetrical reasons, we can easily see that the integral over area b) is the same as over area c).

So overall when $f=\frac{1}{10^i}$:

$$\int_{A_i} fdydx = \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \frac{1}{10^i}dydx + 2\int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{0}^{\frac{1}{2^{i+1}}} \frac{1}{10^i}dydx$$

We'll calculate each integral separately:

$$ \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \frac{1}{10^i}dydx = \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \frac{1}{10^i}*\frac{1}{2^i}-\frac{1}{2}*\frac{1}{10^i}*\frac{1}{2^i}dx =\frac{1}{2} \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}}\frac{1}{20^i}dx=\frac{1}{2}(\frac{1}{20^i}*\frac{1}{2^i}-\frac{1}{2}*\frac{1}{20^i}*\frac{1}{2^i})=\frac{1}{4}*\frac{1}{40^i}$$

Now for the second integral:

$$\int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{0}^{\frac{1}{2^{i+1}}} \frac{1}{10^i}dydx = \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}}\frac{1}{10^i}*\frac{1}{2^{i+1}} dx= \frac{1}{2}\int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}}\frac{1}{20^i}dx=\frac{1}{2}(\frac{1}{20^i}*\frac{1}{2^i}-\frac{1}{2}*\frac{1}{20^i}*\frac{1}{2^i}) = \frac{1}{4}*\frac{1}{40^i} $$

So overall we got:

$$\int_{A_i} fdydx = \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \frac{1}{10^i}dydx + 2\int_{\frac{1}{2^{i+1}}}^{\frac{1}{2^i}} \int_{0}^{\frac{1}{2^{i+1}}} \frac{1}{10^i}dydx=\frac{1}{4}*\frac{1}{40^i}+\frac{1}{2}*\frac{1}{40^i} = \frac{3}{4}*\frac{1}{40^i}$$

And the integral over $Q$:

$$\int_Q fdydx= \int_A f dydx = \sum_{i=0}^{\infty} \int_{A_i} f dydx =\sum_{i=0}^{\infty}\frac{3}{4}*\frac{1}{40^i} = \frac{3}{4}\sum_{i=0}^{\infty}\frac{1}{40^i}$$

According to the formula for sum of converging infinite geometric sequence: $\sum_{i=0}^{\infty}\frac{1}{40^i} =\frac{1}{1-\frac{1}{40}}=\frac{40}{39}$

So my final answer is:

$$\int_Q fdydx= \int_A f dydx = \sum_{i=0}^{\infty} \int_{A_i} f dydx =\sum_{i=0}^{\infty}\frac{3}{4}*\frac{1}{40^i} = \frac{3}{4}\sum_{i=0}^{\infty}\frac{1}{40^i} = \frac{3}{4}*\frac{40}{39} =\frac{10}{13}$$

I do hope that this is the correct answer :)