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

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

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

$$\int\limits_X \big(f(x)-h(x))g(x)d\mu(x)=0$$

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

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

and also for $g_-$ defined similarly.