Are $1/\sqrt{x}$ or $1/x$ Lebesgue integrable on $(0,1)$? If so, why? 
Are $1/\sqrt{x}$ or $1/x$ Lebesgue integrable on $(0,1)$? If so, why and why isn't this true for $1/x$?

I'm having difficulty understanding difference between the above functions in terms of Lebesgue Integration. I'd prefer if the above question could be answered by approximating both functions with simple functions and without using the equality of the Riemann Integral and the Lebesgue Integral.
For example, if we define a simple function like
$$ \phi_n(x) = \lfloor 1/x \rfloor$$
if $1/x \leq n$ and $n$ otherwise, it should be the case that
$$ \int_0^1 \phi_n(x) dx =\infty.$$
But can someone actually show why this is the case?
 A: One rather nice general approach is to note that for any $\varepsilon > 0$, the Riemann integral and Lebesgue integral of these functions coincide on $[\varepsilon,1]$. Now since you have positive functions, the set maps $A \mapsto \int_A f(x) dx$ are actually measures, so you can use continuity of measure to recover the same result that the improper Riemann integral would give, i.e. that $1/\sqrt{x}$ is integrable and $1/x$ is not. 
The same approach works as long as at least one of the positive and negative parts of the function have finite integral. The classic example where this approach doesn't work is $\int_0^\infty \sin(x)/x dx$, which exists in the improper Riemann sense and not in the Lebesgue sense.
I admit that this is overkill, but I think it helps with intuition to get back to something more familiar.
Edit: some clarification follows. The key result here, which is not entirely trivial to prove, is that if $[a,b]$ is a compact interval and $f$ is a (properly) Riemann integrable function on $[a,b]$, then $f$ is a Lebesgue integrable function on $[a,b]$ and $\int_a^b f(x) dx$ (Riemann sense) is the same as $\int_{[a,b]} f(x) dx$ (Lebesgue sense). 
Now Riemann integrable functions are bounded by definition, so the functions in the OP are not Riemann integrable on $[0,1]$ (even after being modified at $x=0$). However they are Riemann integrable on each interval $[\varepsilon,1]$. So we might hope that we could do something to the effect of $\int_{[0,1]} f(x) dx = \lim_{\varepsilon \to 0^+} \int_{[\varepsilon,1]} f(x) dx$, like how we define the improper Riemann integral. 
We can do this provided we can prove that $\int_{[0,1]} f(x) dx = \lim_{n \to \infty} \int_{[1/n,1]} f(x) dx$. (We don't need to worry about different sequences, because $\int_a^1 f(x) dx$ is a decreasing function of $a$.) Now notice that $[1/n,1]=\bigcup_{k=1}^n [1/k,1]$ and $(0,1]=\bigcup_{k=1}^\infty [1/k,1]$. So our equation looks much like the familiar equation for continuity of the measure $\nu$:
$$\nu \left ( \bigcup_{k=1}^\infty A_k \right ) = \lim_{k \to \infty} \nu(A_k)$$
provided $A_k$ is an increasing sequence of sets. In particular, if we can prove that $\mu(A)=\int_A f(x) dx$ is a measure, then we can use continuity of that measure to conclude what we want. The fact that $\mu$ is a measure follows from the monotone convergence theorem.
A: Using simple functions, we can approximate the left half of $$f(x) = 1/x$$ using boxes of width $1/2^n$ and height $2^n$. In particular, say that (draw what this means on paper!)
$$\int_0^\infty 1/x \,\mathrm{d} \mu \geq \sum_{n=0}^\infty 2^n \int_{2^{-n-1}}^{2^{-n}} 1\,d x$$
Then, simplifying the right-hand side gives
$$\sum_{n=0}^\infty 2^n \mu([2^{-n-1}, 2^{-n})) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty $$
so the integral of $f(x) = 1/x$ diverges.
