How may you show that the multivariable integral of a positive function is also positive?

Question

I have a question on the concept of multivariable integration of positive functions, which may be of great interest to those that subscribe to the mathematics stack exchange.

It should be remarked that I have also searched and read previous posts from this mathematics stack exchange. However, these did not help me with the questions to be presented here. Since multivariable calculus is a great topic here, I believe these questions, along with their possible discussions and answers, may be of great interest to those that subscribe to the mathematics stack exchange.

With that said, it is well-known that if a function $f(x)$ is non-negative (non-postive), then its integral is non-negative (non-positive). More precisely, let $f:[a,b]\rightarrow \mathbb{R}$ be an integrable positive function, then $\int^{a}_{b}f(x)dx>0$.

With this result in mind, suppose that we have the following integral

$$I=\int \int \left(\nabla\times\vec{F}\right)\cdot d\vec{A}=\int \int dxdy\left(\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\right).$$ Based on the above, I ask:

1. If

$$\left(\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\right)>0$$,

would that imply that $I>0$?

2. Is there any reference in which this result is discussed or demonstrated?

Answer 1

Only if the set you are integrating over has nonzero area. For example if $F(\boldsymbol x)>0$ then $$\int\limits_{|\boldsymbol x|<1}F(\boldsymbol x)\mathrm d^2\boldsymbol x>0$$ Because the set $$\{\boldsymbol x\in\Bbb R^2:|\boldsymbol x|<1\}$$ Has positive area. On the other hand, $$\int\limits_{x_1\in[0,1]~,~x_2=0}F(\boldsymbol x)\mathrm d^2\boldsymbol x=0\tag{1}$$ Since the set $$\{(x_1,x_2)\in\Bbb R^2:x_1\in[0,1]~\text{and}~x_2=0\}$$ Has zero area.

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This essentially comes down to how you measure sets. If considered the same integral as equation $(1)$, but integrated with respect to the measure $\mathrm dx_1$ instead of the measure $\mathrm d^2\boldsymbol x$ we would get $$\int\limits_{x_1\in[0,1]~,~x_2=0}F(x_1,x_2)\mathrm dx_1 \\ =\int_0^1 F(x_1,0)\mathrm dx_1>0$$ Because now the set $$\{(x_1,x_2)\in\Bbb R^2:x_1\in[0,1]~\text{and}~x_2=0\}$$ Has positive measure with respect to the measure $\mathrm dx_1$.

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Generally speaking, let $(X,\Sigma,\mu)$ be a measure space and let $f:X\to\Bbb R_+$ be integrable and everywhere positive. Let $E\in\Sigma$ be a measurable set. The following hold:

• If $\mu(E)=0$, then $\int_E f~\mathrm d\mu=0$.
• If $\mu(E)\neq 0$, then $\int_E f~\mathrm d\mu >0$.

See for instance this MSE post for proof.