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Chapter 11: Problem 12

Show that if \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and \(f \geq g\) a.c., then $$ \int_{E} f d \mu \geq \int_{E} g d \mu $$

Short Answer

Expert verified
Integrals preserve order. That is, if \(f \geq g\) almost everywhere, then the integral of \(f\) over \(E\), \(\int_{E} f d \mu\), is greater than or equal to the integral of \(g\) over \(E\), \(\int_{E} g d \mu\).

Step by step solution

01

Establish the condition

We are given \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and that \(f \geq g\) almost everywhere (a.c.) in \(E\). This means that the set \({x \in E: f(x) < g(x)}\) has measure zero. Since \(f\) and \(g\) are Lebesgue integrable, we can subtract them without issue.
02

Formulate the measure of difference inequality

We define the difference \(h\) by \(h = f - g\). Because \(f \geq g\) almost everywhere, \(h(x) \geq 0\) for almost all \(x\) in \(E\). By the non-negativity property of Lebesgue integrals, which states that for any non-negative function \(h\), \(int_E h d \mu \geq 0\). In this case, the inequality becomes, \(int_E f d \mu - int_E g d \mu = int_E (f - g) d \mu = int_E h d \mu \geq 0\).
03

Rearrange inequality to show condition

Re-arrange the inequality obtained in the previous step to obtain \(int_E f d \mu \geq int_E g d \mu\). This shows that if \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and \(f \geq g\) almost everywhere, then the integral of \(f\) over \(E\) will be greater than or equal to the integral of \(g\) over \(E\).

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Most popular questions from this chapter

If \(E\) is bounded and there exists an interval \(I \supset E\) such that $$ \mu^{*}(I)=\mu^{*}(I \cap E)+\mu^{*}(I-E) $$ then this holds for all intervals, possibly even those overlapping \(E\).

Find a step function \(s(x)\) that approximates \(f(x)=x^{2}\) uniformly to within \(\varepsilon>0\) on \([0,1]\), in the sense that \(|f(x)-s(x)|<\varepsilon\) cierywhere in \([0,1]\).

Let \(f: X \rightarrow \mathbb{R}\) and \(g: X \rightarrow \mathbb{R}\) be mcasurable functicns and \(E \subset X\) a measurable set. Show that $$ h(x)= \begin{cases}f(x) & \text { if } x \in E \\ g(x) & \text { if } x \not E\end{cases} $$ is a measurable function on \(X\).

If \(\mu^{*}(N)=0\) show that for any set \(E, \mu^{*}(E \cup N)=\mu^{*}(E-N)=\mu^{*}(E)\). Hence show that \(E \cup N\) and \(E-N\) are Lebesgue measurable if and only if \(E\) is measurable.

\(\operatorname{lf}(X, M)\) and \((Y, N)\) are measurable spaces, show that the projection maps \(\mathrm{pr}_{1}: X \times\) \(Y \rightarrow X\) and \(\mathrm{pr}_{2}: X \times Y \rightarrow Y\) defined by \(\mathrm{pr}_{1}(x, y)=x\) and \(\mathrm{pr}_{2}(x, y)=y\) are measurable functions,
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