I am studying real analysis from the book which is written by Manfred Stoll and somewhere in Lebesgue integral properities(page:476) I stuck. I don't believe totally prove the property so I will be happy if you could tell me where i did wrong. Thanks in advance


$A \subset \mathbb{R}$ is a measurable bounded set and $f,g :A \rightarrow \mathbb{R}$ are measurable and non-negative functions. If $f \leq g$ $ \,$a.e. on the set A then

$\int_A f d \lambda \leq \int_A g d \lambda$ $\quad$ holds.

My attempt:

I have the definition about Lebesgue integrability and (proved) lemma which i used for proof are in below;

Definition : Let $A \subset \mathbb{R}$ bounded and measurable set.

$f:A \rightarrow \mathbb{R}$ non-negative real valued function.

Define , $f_n(x)= min\{f(x),n\}$ then,

$\int_A f d \lambda = lim_{n \rightarrow \infty} \int_A f_n d\lambda$

Lemma*: $ f: [a,b] \rightarrow \mathbb{R}$ is bounded and measurable function and $f \leq g$ a.e. on [a,b] then

$\int_{[a,b]} f d \lambda \leq \int_{[a,b]} g d \lambda$

Proof(of the proposition): Let $ f \leq g$ $\,$ a.e on the set bounded A. Then there exits $a,b \in \mathbb{R}$ such that $A \subset [a,b]$. We can define $g_{x_{A}}$ and $f_{x_{A}}$ as ;

$f_{x_{A}}= \begin{cases} f(x) & x \in A\\ 0 & x \notin A \end{cases}$


$g_{x_{A}}= \begin{cases} g(x) & x \in A\\ 0 & x \notin A \end{cases}$

Now we will define the function sequences

$f_n ,g_n :[a,b] \rightarrow \mathbb{R}$ as;

$f_n(x)= min\{f(x),n\}$ $\quad $ and $\quad$ $g_n(x)= min\{g(x),n\}$

So by using Lemma*, we can say that $f_n$ and $g_n$ function sequences are measurable $ \forall n \in \mathbb{N}$ and $ f_n \leq g_n$ $\,$ a.e on [a,b] so that

$\int_{[a,b]} f_n d \lambda \leq \int_{[a,b]} g_n d \lambda$ $\iff$ $\int_A f_n d \lambda \leq \int_A g_n d \lambda$

Now take limit from both sides of inequility,

$\int_A f d \lambda = lim_{n \to \infty}\int_A f_n d \lambda \leq lim_{n \to \infty}\int_A g_n d \lambda =\int_A g d \lambda$

  • $\begingroup$ What exactly is Your question? $\endgroup$ – Peter Melech Apr 15 '18 at 11:42
  • $\begingroup$ I am trying to prove the proposition $\endgroup$ – Alper Erdem Apr 15 '18 at 11:43
  • $\begingroup$ You are making things too complicated. The hypothesis tells you that $I_A f \leq I_A g$ everywhere. Just integrate both sides. $\endgroup$ – Kavi Rama Murthy Apr 15 '18 at 11:50
  • $\begingroup$ But the OP wants to show the monotonicity of the integral, just "integrating both sides" would mean to use it, besides @Alper Erdem don´t You mean $f_n(x)=\min\{f_{\chi_A}(x),n\}$ and similary for $g_n$. Did You prove the Lemma? $\endgroup$ – Peter Melech Apr 15 '18 at 11:54
  • $\begingroup$ Couldn't I integrate both sides? I know for the function sequences $f_n$ and $g_n$ are both monotone increasing and converges either to a real number or diverges to $\infty$ . Yes i have proved the lemma $\endgroup$ – Alper Erdem Apr 15 '18 at 12:00

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