Measure theory question on integrals.

[bolzano]

Hi, I was wondering whether if ∫f×g dμ=∫h×g dμ for all integrable functions g implies that f = h?

Thanks

 

[mathman]

f = h almost everywhere, assuming the integrals exist.

 

[mathwonk]

equality of integrals never implies equality, only equality a.e.

 

[bolzano]

Hi, thanks for the replies. However is this something trivial or is it hard to prove (that thery're equal a.e.)?

 

[jostpuur]

Your claim was not properly explained, but I believe it is trivial to reduce it to this claim:

If \int\limits_X f(x)d\mu(x)=0 and f(x)\geq 0 for all x\in X, then f(x)=0 for \mu-"almost all" x\in X.

This claim is not trivial. You must use the properties of measures and integrals.

Assume that there exists a set A\subset X such that f(x)>0 for all x\in A, and also \mu(A)>0. Now you can define sets

<br /> A_1 = \{x\in A\;|\; f(x)&gt;1\}<br />
<br /> A_n = \Big\{x\in A\;\Big|\; \frac{1}{n-1}\geq f(x) &gt; \frac{1}{n}\Big\},\quad\quad n=2,3,4,\ldots<br />

Equality \mu(A)=\sum_{n=1}^{\infty}\mu(A_n) will imply that at least one of the \mu(A_n) is positive.

 

[jostpuur]

If

<br /> \int\limits_X \big(f(x)-h(x))g(x)d\mu(x)=0<br />

holds for all integrable g, then it will also hold for g_+ defined by

<br /> g_+(x)=\left\{\begin{array}{ll}<br /> 1,\quad &amp;f(x)-h(x)&gt;0\\<br /> 0,\quad &amp;f(x)-h(x)\leq 0\\<br /> \end{array}\right.<br />

and also for g_- defined similarly.