In this Lebesgue integral problem, I was thinking as an alternative to passing the limit under the integral,
- Let $u=1+nx^2$. Then
$$\lim_{n\to\infty}\int_{[0,1]}\frac{nx}{1+nx^2} = \lim_{n\to\infty}\int_{[1,n+1]}\frac{\frac{1}{2}}{u} = \lim_{n\to\infty} \frac{1}{2} \ln|u| \mid_{[1,n+1]} = \lim_{n\to\infty} \frac{1}{2} \ln|n+1| = \infty$$
Can we not substitute for Lebesgue integrals as do we Riemann integrals?
- Say $\lim_{n\to\infty}\int_{[0,1]}\frac{nx}{1+nx^2} = \lim_{n\to\infty}\int_0^1 \frac{nx}{1+nx^2}dx$ and then compute the latter.
Or can we say such only if we compute the limits in the first place?
- Say $\int_{[0,1]}\frac{nx}{1+nx^2} = \int_0^1 \frac{nx}{1+nx^2}dx$ and then compute the latter.
Is it wrong to say that $\frac{nx}{1+nx^2}$ is Riemann integrable on $[0,1]$ and thus is Lebesgue integrable?
I guess Lebesgue integrals don't follow the same rules as Riemann integrals, so somewhere there's an illegal step. Which please?
1) In the original problem, the limit is outside the integral. So if you want to exchange limit and integral, you have to cite applicable theorems under which condition that is possible (and then check these conditions). I'll come back to this later.
2,3) Since the limit is outside the integral, any statement using any integral definition (Rieman or Lebesgue) only makes sense if the integral exist. Since Riemann integrability ensures Lebesgue integrability, you can solve the integral part first, and in this case is turns out the function is Rieman integrable, so you can use whatever method to calculate that integral.
You wrote
You seem to be under some misconception that something you did is wrong, but I don't see anything wrong.
4) As one can easily see, the integrand converges pointwise to $f(x)=0$ on $[0,1]$ for $n \to \infty$, so exchanging limit and integral would lead to quite an incorrect result. The main reason is that the convergence is not uniform, for each $n$ there are arguments $(\frac1{\sqrt{n}})$ where the integrand is quite far away from 0!.