Properties of Lebesgue Integral

[nateHI]

Homework Statement

Let $f,g$ be non-negative Lebesgue measurable functions, and let $E$ be a measurable set. Prove that:

$f\ge g\implies \int_E f d\lambda \le \int_E g d\lambda$

Homework Equations

The Attempt at a Solution

I've been trying to apply Fatou's lemma but haven't been getting anywhere.

 

[nateHI]

Homework Statement

Let $f,g$ be non-negative Lebesgue measurable functions, and let $E$ be a measurable set. Prove that:

$f\ge g\implies \int_E f d\lambda \le \int_E g d\lambda$

Homework Equations

The Attempt at a Solution

I've been trying to apply Fatou's lemma but haven't been getting anywhere.

Here's a little more detail on my attempt
$\int (f-g)\le liminf_{n\to\infty}\int{(f_n-g_n)}$
$\implies \int{f}-\int{g}\le liminf_{n\to\infty}\int{f_n}-liminf_{n\to\infty}\int{g_n}$
$\implies -\int g -limsup_{n\to \infty}\int{-g_n} \le -\int f -limsup_{n\to \infty}\int{-f_n}$
$\implies \int g +limsup_{n\to \infty}\int{g_n} \ge \int f +limsup_{n\to \infty}\int{f_n}$
$\implies \int g \ge \int f$

My confusion arises when I compare this to monotonicity. I must be missing a step and it must be because I'm not using the fact that
$\int_E f=\int{f \mathcal{X}_E}$

 

[Ray Vickson]

Homework Statement

Let $f,g$ be non-negative Lebesgue measurable functions, and let $E$ be a measurable set. Prove that:

$f\ge g\implies \int_E f d\lambda \le \int_E g d\lambda$

Homework Equations

The Attempt at a Solution

I've been trying to apply Fatou's lemma but haven't been getting anywhere.

You are getting nowhere because the result is false. Try the example $f(x) = 2, g(x) = 1 \; \forall x$ and $E = [1,2]$.

 

[Fredrik]

I think one of those inequalities is supposed to point in the other direction. For example, if $f\leq g$, then $\int_E f\mathrm d\lambda\leq \int_E g\mathrm d\lambda$. There's more than one definition of the Lebesgue integral, so you will have to tell us which one you're using. Is $\lambda$ the Lebesgue measure on $\mathbb R^n$, or just some arbitrary measure?

Regardless of what definition you're using, it will probably be slightly easier to break this up into two parts:
(a) If $f\geq 0$ then $\int_E f\mathrm d\lambda\geq 0$.
(b) If $f\leq g$ then $\int_E f\mathrm d\lambda\leq \int_E g\mathrm d\lambda$.

(a) is slightly easier to prove than (b) alone, and if you have proved (a), it's very easy to use it to prove (b).

 

[nateHI]

You are getting nowhere because the result is false. Try the example $f(x) = 2, g(x) = 1 \; \forall x$ and $E = [1,2]$.

Ack! That's not good, this was the professors practice problem for the upcoming test. Also, I double checked and the problem is stated here as he gave it to the class. Please disregard my request for assistance.

 

[Fredrik]

Then there's a typo in the problem. If you reverse the direction of one of the inequalities, you get one of the standard theorems that's proved in every book on integration.