Proofing the identity of double integrals over symmetric function

Question

In one of my physics classes we had given a double integral over a function $f$ in two variables: $$G=\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x$$

The function $f$ is symmetric in its parameters: $$f(x,y)=f(y,x)$$

We used this property to simplify the integral for later numerical integration: $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=2\int_{x\in D}\int_{y\in D, y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x$$

Although this makes intuitive sense to me, I was unable to prove it using the given equations alone. How can I properly prove that the equality given above is true?

Attempt

My idea for a proof so far was to do the following expansion: $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{x\in D}\int_{y\in D,y\ge x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x$$

Renaming $x\rightarrow y'$ and $y\rightarrow x'$ in the second integral: $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{y'\in D}\int_{x'\in D,x'\ge y'}f(y',x')\ \mathrm{d}x'\ \mathrm{d}y'$$

Using the symmetry and reversing the condition $x'\ge y'$ we get halfway there: $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{y'\in D}\int_{x'\in D,y'\le x'}f(x',y')\ \mathrm{d}x'\ \mathrm{d}y'$$

However, the order of integration is still strong and I do not know how to fix this. What am I missing?

Answer 1

Taking my attempt so far and the hints I got in the comments, the proof goes as follows:

Proof

1. Expand the integral. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{x\in D}\int_{y\in D,y\ge x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x$$

2. Rename $x\rightarrow y'$ and $y\rightarrow x'$ in the second integral. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{y'\in D}\int_{x'\in D,x'\ge y'}f(y',x')\ \mathrm{d}x'\ \mathrm{d}y'$$

3. Use the symmetry $f(x,y)=f(y,x)$) and reverse the condition $x'\ge y'$. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{y'\in D}\int_{x'\in D,y'\le x'}f(x',y')\ \mathrm{d}x'\ \mathrm{d}y'$$

4. Replace the boundary conditions with the step function $\chi(x,y)$. $$\chi(x,y)=\begin{cases}1 & \text{if}\ y\le x\\0 & \text{otherwise}\end{cases}$$ $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D}f(x,y)\chi(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{y'\in D}\int_{x'\in D}f(x',y')\chi(x',y')\ \mathrm{d}x'\ \mathrm{d}y'$$

5. Apply Fubini's theorem. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=\int_{x\in D}\int_{y\in D}f(x,y)\chi(x,y)\ \mathrm{d}y\ \mathrm{d}x+\int_{x'\in D}\int_{y'\in D}f(x',y')\chi(x',y')\ \mathrm{d}y'\ \mathrm{d}x'$$

6. Simplify. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=2\int_{x\in D}\int_{y\in D}f(x,y)\chi(x,y)\ \mathrm{d}y\ \mathrm{d}x$$

7. Put $\chi(x,y)$ back as a boundary condition. $$\int_{x\in D}\int_{y\in D}f(x,y)\ \mathrm{d}y\ \mathrm{d}x=2\int_{x\in D}\int_{y\in D,y\le x}f(x,y)\ \mathrm{d}y\ \mathrm{d}x\ \blacksquare$$

Note

Step 5. applies Fubini's theorem, and thus implicitly implies: $$\int_{D\times D}|f(x,y)\chi(x,y)|\ \mathrm{d}(x,y)<+\infty\text{,}$$ which in turn implies: $$\int_{D\times D}|f(x,y)|\ \mathrm{d}(x,y)<+\infty\text{.}$$