I know that I can derive the CDF of $Y = \mathrm{min}\left(X_1, X_2\right)$ if $F(x)$ is the CDF of $X_1$ and $X_2$ (which are IID) as
$$P(Y<y) = 1-(1-F(y))^2 = 2 F(y) - F(y)^2$$
Which would then for the pdf of $Y$ give $$f_{Y}(y) = 2 f(y) - 2 F(y)f(y)$$
Where $f(x)$ is the pdf of $X$. I'm trying to arrive at the same result by using the following approach for change of variables for PDFs.
$$f_{Y}(y) = \int_{\mathbb{R}^2}f(x_1)f(x_2)\delta(y - g(x_1, x_2 ))\mathrm{d}x_1\mathrm{d}x_2$$
where $g(x_1, x_2)$ is the transformation that moves $(x_1,x_2) \rightarrow y$. In our case
$$g(x_1, x_2) = \mathrm{min}(x_1,x_2)$$
I proceeded like so (below $I$ is the indicator function)
\begin{align}f_{Y}(y) &= \int_{\mathbb{R}^2}f(x_1)f(x_2)\delta\left(y - \mathrm{min}(x_1,x_2)\right)\mathrm{d}x_1\mathrm{d}x_2 \\ &= \int_{\mathbb{R}^2}f(x_1)f(x_2)\delta\left(y - I(x_1 < x_2)x_1 - I(x_1 > x_2 )x_2\right)\mathrm{d}x_1\mathrm{d}x_2 \\ &= \int_{\mathbb{R}}f(x_2)\left(\int_{-\infty}^{x_2}f(x_1)\delta(y-x_1)\mathrm{d}x_1 + \int_{x_2}^\infty f(x_1)\delta(y-x_2)\mathrm{d}x_1\right)\mathrm{d}x_2\\ &= \int_{\mathbb{R}} f(x_2) f(y)\mathrm{d}x_2 + \int_{\mathbb{R}}f(x_2)\left(1 - F(x_2)\right)\delta(y-x_2)\mathrm{d}x_2\\ &= f(y) + f(y)(1-F(y))\\ &= 2 f(y) - f(y)F(y) \end{align}
So I'm missing a factor of 2 from the second term. Where did I go wrong?
This step: \begin{align} &= \int_{\mathbb{R}}f(x_2)\left(\int_{-\infty}^{x_2}f(x_1)\delta(y-x_1)\mathrm{d}x_1 + \cdots\right)\mathrm{d}x_2\\ &= \int_{\mathbb{R}} f(x_2) f(y)\mathrm{d}x_2 + \cdots \end{align} is only correct when the $\delta$ is the the range of integration, i.e. $y<x_2.$ That term should be $\int_{y}^{\infty} f(x_2) f(y)\mathrm{d}x_2,$ in other words $f(y)(1-F(y)).$