0
votes
0answers
24 views

Can the theory of Lebesgue integration be extended in a way analogous to extending Riemann integrals to improper Riemann integrals?

I recently (last night) learned the definition of Lebesgue integration and one of the limitations I was told was that some improper Riemann integrals aren't Lebesgue integrable. It occurred to me ...
3
votes
1answer
43 views

Conditions on a measure

Suppose we have a measure $\mu$ on $\mathbb R_+$ such that $\forall s>0$ $t^s\in L^1(\mathrm d\mu(t))$, but functions $\mathbf{1}_{t>0} $ and $\mathbf{1}_{t\in (0,1)}$ are not necessarily in ...
5
votes
3answers
187 views

$ \int^{\infty}_0 |\frac{1}{(1+x)\sqrt x}|^p ~ \mathrm dx < \infty \implies p=?$

If $ f(x) = \frac{1}{(1+x)\sqrt x} $ how to find all $ p > 0 $ such that $$ \int^{\infty}_0 |f(x)|^p dx < \infty $$ The integral is with respect to lebesgue measure. Any solution or hints would ...
0
votes
1answer
92 views

The relation between arbitrary measure space and the Lebesgue integral

Let $(X, \mathcal F, \mu)$ be a measure space and $f\in M^+(X,\mu)$ (the measurable non-negative functions), and $t>0$. Now let $$S_f(t)=\{x\in X:f(x)>t\} \quad \Psi_f(t)=\mu(S_f(t))$$ Prove ...
3
votes
1answer
129 views

the integral of a periodic function

Consider $f:\mathbb{R}\to \mathbb{C}$ a bounded and $1$-periodic function, and $g \in L^1(R)$ then $$\lim_{n\to \infty} \int _{R}g(x)f(nx)dx=\int_0^1f(s)ds\int_R g(t)dt.$$ I think the fact that $f$ ...
3
votes
2answers
178 views

Lebesgue Integral, existence, improper integrals, etc.

Problem: At the request of another user, I am taking an older question and specifically addressing one problem. I am self-learning about Lebesgue integration, and am just starting to try and apply ...
30
votes
7answers
2k views

Why do we restrict the definition of Lebesgue Integrability?

The function $f(x) = \sin(x)/x$ is Riemann Integrable from $0$ to $\infty$, but it is not Lebesgue Integrable on that same interval. (Note, it is not absolutely Riemann Integrable.) Why is it we ...