1. ## Discontinuity and integration

Let $f(x) = x^{-1/2}$ if 0 < x < 1, $f(x) = 0$ otherwise. Let $\{r_n\}$ be an enumeration of rationals, and define $g(x) = \sum (1/2^n)f(x-r_n)$. Show that g is discontinuous at every point and unbounded on every interval.

Show that if f is a non-negative integrable function, for every $\epsilon > 0$ there exists $E \in M$ such that $m(E) < \infty$ and $\int_{E} f > \int f - \epsilon$

2. Originally Posted by davidmccormick
Let $f(x) = x^{-1/2}$ if 0 < x < 1, $f(x) = 0$ otherwise. Let $\{r_n\}$ be an enumeration of rationals, and define $g(x) = \sum (1/2^n)f(x-r_n)$. Show that g is discontinuous at every point and unbounded on every interval.

Show that if f is a non-negative integrable function, for every $\epsilon > 0$ there exists $E \in M$ such that $m(E) < \infty$ and $\int_{E} f > \int f - \epsilon$
I have no clue about the first one

For question #2:

I'm not sure what M is? Is that the domain of the function? Assuming it is

Let $S_L$ = {x \in M: f(x) > L }

Notice that $L*\chi_{S_L} \leq f(x)$ and thus $\int L*\chi_{S_L} \leq \int f \implies Lm^*(S_L) \leq \int f \implies m^*(S_L) \leq \frac{\int f}{L}$

Since f is integrable, the numerator is bounded and it follows that as L tends to infinity, $m^*(S_L)$ tends to 0.

I'll leave it up to you to show that $\int_{S_L} f \rightarrow 0$ as $L \rightarrow \infty$ and so we can find an L large enough such that $\int_{S_L} f < \epsilon$ and so the set we are looking for is E = M \ $S_L$

Note, this used the fact that we can break the integral up into disjoint sets.

3. Well, since every interval contains a rational, and g blows up at at every rational, g is unbounded in every interval. No idea for discontinuity. That is an absolutely horrible function. Is its graph dense in $\mathbb{R}^2$?