The function $g$ is a composition of functions as follows ($\chi_A$ is the characteristic function of the set $A$):
$$f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$$
For an enumeration of the rationals $\{r_i\}$: $$g(x) = \sum_{n=1}^\infty 2^{-n}f(x-r_n)$$
Prove that $g \in L^1(m)$ where $m$ is the Lebesgue measure for $m$ almost everywhere on $x$. Prove that $g$ is unbounded on every open interval $I \in [0,1]$ and that $\forall I, \forall N \in \mathbb{N} \Rightarrow m(\{x\in I : g(x)>N\}) > 0$
Putting everything in place, we have
$$\int \sum_{n=1}^\infty 2^{-n}\frac{\chi_{[0,1](x-r_n)}}{\sqrt{x-r_n}} dm$$
Analyzing the parts of this function, we see that $2^{-n} \geq 0; \chi \geq 0$ I don't know how to show that with the denominator this is a function in $L^+$. I know that for being in $L^1$ I am interested in $|f|$ but I am still unsure if I should and how to proceed about this.
If I could show that, then we can swap the sum and the integral and get
$$\sum_{n=1}^\infty \int 2^{-n}\frac{\chi_{[0,1](x-r_n)}}{\sqrt{x-r_n}} dm$$
where
$$\int 2^{-n}\frac{\chi_{[0,1](x-r_n)}}{\sqrt{x-r_n}} dm \leq \int 2^{-n}\frac{1}{\sqrt{x-r_n}} dm \leq \int \frac{1}{\sqrt{x-r_n}}dm$$
Do I proceed with some sort of limit theorem or keep trying to prove that this is less than some other function?
Monotone convergence theorem (@GNU, @zhw in the comments) with translation invariance of the Lebesgue measure implies that $$ \int_\Bbb R \sum_{n\ge 1}2^{-n}f(x-r_n)\mathrm dm = \sum_{n\ge 1}2^{-n} \int_\Bbb Rf(x-r_n)\mathrm dm =\sum_{n\ge 1}2^{-n} \int_\Bbb R f=\int_0^1 \frac{\mathrm dx}{\sqrt{x}}=2. $$ So, $g\ge 0$ is in $L^1(\Bbb R,m)$. To see that $g$ is unbounded on any $I$ and $m\{x\in I:g(x)>N\}>0$ for all $N$, note that there is $r_k\in I$ for some $k$. Also note that since $I$ is open, we can choose $\delta$ so that $(r_k,r_k+\delta)\subset I$ and at the same time $\delta<\min\{\frac1{4^kN^2},1\}$. If $x\in (r_k,r_k+\delta)$, then we have $$ g(x)\ge 2^{-k}f(x-r_k)=2^{-k}\frac1{\sqrt{x-r_k}}\ge 2^{-k}\frac1{\sqrt\delta}>2^{-k}\cdot 2^kN =N. $$ We can see that $$m\{x\in I:g(x)>N\}\ge m(r_k,r_k+\delta)=\delta>0$$ and since $N$ is arbitrary, $g(x)$ is unbounded as $x\xrightarrow{} r_k^+$.