Does a piecewise rational/irrational defined function have a bound on its integral? 
$$f(x)= \left\{\begin{array}{ll}
x/3 &\mbox{if $x$ is rational,}\\
x/2 &\mbox{if $x$ is irrational.}
\end{array}\right.$$

Suppose that $f$ is my function, can I say that the integral of this function between $0$ and any positive number is greater than the integral of $x/3$? I want a good reason why I can't put a lower bound on this integral if the answer is no, since it clearly appears to have an upper and lower bound to it.
 A: 
I want a good reason why I can't put a lower bound on this integral if the answer is no, since it clearly appears to have an upper and lower bound to it.

The reason is that there is no such (Riemann) integral. Since every interval has both rational and irrational numbers in it, we know
\begin{align*}
    U(f,P) &= U(x/2,P) \\
    L(f,P) &= L(x/3,P)
\end{align*}
for every partition $P$ of an interval $[a,b]$.  Therefore,
\begin{align*}
    \inf_P U(f,P) = \inf_P U(x/2,P) = \int_a^b (x/3)\,dx \\
    \sup_P L(f,P) = \sup_P L(x/3,P) = \int_a^b (x/2)\,dx
\end{align*}
Since the two on the right are different, $f$ is not integrable on $[a,b]$.  
@sranthrop points out that there is a different answer when using the Lebesgue integral.  In that theory we can partition sets into subsets that are not intervals.  For instance, the interval $[a,b]$ can be partitioned into the rational elements (call this set $E$) and irrational elements ($E'$).  Since $E$ is of measure zero,
$$
\int\limits_{[a,b]} f\,d\mu = \int\limits_E f\,d\mu + \int\limits_{E'} f\,d\mu 
= \int\limits_{E'} f\,d\mu = \int_a^b (x/2)\,dx
$$
