What is the Lebesgue integral of $1/x$ on $[0,1]$?

Question

I've seen this question asked a couple times here but the answers use materials I haven't learned yet. I think I should be able to solve this by finding $f_n(x)$ that converges to $f(x)$, but I don't really know any further than that. Any suggestions would be helpful. I should be able to solve this without invoking the Lebesgue measure.

Answer 1

It's infinite; you can do that with Riemann sums, dividing the interval into $n$ equal subintervals. The integral on the interval $[\frac{m}{n}, \frac{m+1}{n}]$ is between $\frac{1}{n} \times \frac{n}{m+1}$ and $\frac{1}{n} \times \frac{n}{m}$ (i.e. "width of subinterval" times "min/max value on the subinterval"); i.e. between $\frac{1}{m+1}$ and $\frac{1}{m}$, and in particular at least $\frac{1}{m+1}$. So the total Riemann sum is at least $\sum_{m=0}^{n-1} \frac{1}{m+1}$ for every $n$.

Answer 2

$f(x)=1/x$ is not $L^1([0,1])$. Consider the Riemman integral of $g_{\epsilon}:=f1_{[\epsilon,1]}$ on $[0,1]$, which gives $$ \int_{\epsilon}^1f(x)\ dx=\log x\big|_\epsilon^1=-\log\epsilon $$ Now let $\epsilon\to0+$.