I cannot understand the following passage in my notes. Seems like something very trivial (or maybe not) but I'm stuck.
For every $f, g$ in $L^2([0, 1])$ we have $$\int_0^1 \text{d}x\int_0^x \text{d}t \bar{f(x)} g(t) = \int_0^1\text{d}t \int_t^1\text{d}x \bar{f(x)} g(t)$$
Where the bar denotes the complex conjugate. I don't understand the passage from the integral $\int_0^x$ to $\int_t^1$.
You're integrating over the region $$R=\{(x,t)\in [0,1]^2: 0\leqslant x\leqslant 1,0\leqslant t\leqslant x\}.$$ This region admits the alternative description $$\{(x,t)\in [0,1]^2:0\leqslant t\leqslant 1, t\leqslant x\leqslant 1\}.$$
This is most easily seen by drawing the square $[0,1]^2$ and shading the region. Think of a horizontal $x$ axis and vertical $t$ axis. Draw the diagonal line $x=t$. Then the region $R$ is the triangular region below (or, equivalently, to the right of) the line. You can first note that $x$ will range from $0$ to $1$ in the region. Given an $x$, $t$ will range from $0$ to $x$ (just check the shaded part on a given vertical line at a given $x$). Going in the other order, $t$ will range from $0$ to $1$ and, given a $t$, $x$ will range from $t$ to $1$ (again, checking the shaded part on a given horizontal line at a given $t$).
You're using Fubini's theorem to switch the order of integration, which is valid, since $\overline{f}, g\in L_2$, so $\int\int |f(x)g(t)|dtdx<\infty$.