Using the techniques that I have been taught, I have attempted to prove the following result:
$$\int_a^b \int_c^d f(x)g(y) \space dydx = \left(\int_a^b f(x) dx\right)\left(\int_c^d g(y) dy\right)$$
I am aware of that there is already a MathStackExchange post with solutions to this problem, however, this post is specifically dedicated to a solution verification and therefore it is not a duplicate of the linked post.
I would like a review of my alternative attempt to those in the linked post. It is different to the solutions presented there and I am wondering if this is a valid attempt:
\begin{aligned}
\int_a^b \int_c^d f(x)g(y) \, \mathrm{d}y\,\mathrm{d}x &= \lim_{\Delta x \to 0+} \lim_{\Delta y \to 0+} \sum_{i = 1}^{\Big\lceil\frac{b – a}{\Delta x}\Big\rceil} \sum_{j = 1}^{\Big\lceil\frac{d – c}{\Delta y}\Big\rceil} f(a + i\Delta x)g(c + j\Delta y)\Delta x\Delta y \\
&= \Bigg( \lim_{\Delta x\to 0+} \sum_{i = 1}^{\Big\lceil \frac{b – a}{\Delta x} \Big\rceil} f(a + i \Delta x)\Delta x \Bigg) \Bigg( \lim_{\Delta y \to 0+} \sum_{j = 1}^{\Big\lceil \frac{d – c}{\Delta y} \Big\rceil} g(c + j\Delta y) \Delta y\Bigg) \\
&= \Bigg( \int_a^b f(x) \, \mathrm{d}x \Bigg) \Bigg( \int_c^d g(y) \, \mathrm{d} y \Bigg).
\end{aligned}
More explicitly, my question is whether or not the above proof attempt is correct, and if so, is this the best way to prove the desired result?
Best Answer
What you have done is correct, but overcomplicated.
Linearity of the integral tells us that the following holds:
$$ \int_a^b \big{(}\alpha f(x) + \beta g(x) \big{)} \space dx = \Big{(}\alpha \int_a^b f(x) \space dx \Big{)}+ \Big{(}\beta \int_a^b g(x)\space dx \Big{)}$$
The above relies on the fact that $\alpha , \beta$ are fixed with respect to $x$. However, we can extend this to the fact that $f(x)$ is fixed with respect to $y$. Therefore, when we look at the term $f(x)g(y)$ within an integral with respect to $y$, we can treat $f(x)$ as if it as a constant.
Therefore we can simply use linearity (as above) to justify the following:
$$ \color{red}{\int _c^d f(x)g(y) \space dy} = f(x) \int _c^d g(y) \space dy$$
Now you can just substitute this into the expression you have involving the double integral and it will give you the desired result without needing to use Riemann sums (which is not necessary for this problem).
We have:
$$\int_a^b \color{red}{\int_c^d f(x)g(y) \space dy} \space dx $$
But because we have already simplified the $\color{red}{\text{red}}$ part of this formula above (using linearity), we can replace it with the simplified expression to give us the desired result.