Orthogonal complement to symmetric functions in $L^2$ Given $$M=\{ f\in L^2([-1,1])\; | \; f(x)=f(-x) \text{ for almost every } x\in [0,1]\}$$
I have already proved that $M$ is a closed subspace of $L^2([-1,1])$, but have yet to find $M^\bot$.
Obviously, $\int_{[-1,1]}g\;dx=0$ for all $g\in M^\bot$ because choose $f=1\in M$.
It seems to me that $M^\bot$ might be $\{g\in L^2([-1,1])\; | \; g(x)=-g(x) \text{ for almost every } x\in[0,1]\}$, but I do not know how to proceed.
 A: First we show $$\{g\in L^2([-1,1])\; | \; g(-x)=-g(x) \text{ for almost every } x\in[0,1]\}\subset M^\bot.$$
Proof. Let $g(-x)=-g(x)$ for almost every $x\in[0,1]$ and $f\in M$, define $h(x)=f(x)\cdot g(x)$, then $$h(-x)=f(-x)\cdot g(-x)=f(x)\cdot (-g(x))=-h(x). $$ for almost every $x\in[0,1]$. Substituting $y=-x$ and $\frac{dy}{dx}=-1$ at $(*)$ leads to
\begin{align*}\int_{-1}^1f(x)g(x)\ dx&=\int_{-1}^1h(x)\ dx\overset{(*)}{=}\int_1^{-1}h(-y)\cdot (-1)dy\\&= \int_{-1}^{1}h(-y)\ dy=\int_{-1}^1(-h(x))\ dx\\&=-\int_{-1}^1f(x)g(x)\ dx.\end{align*}
Hence, $\int_{-1}^1f(x)g(x)\ dx=0$ and $g\in M^\bot$.
Now, we show
$$M^\bot\subset\{g\in L^2([-1,1])\; | \; g(-x)=-g(x) \text{ for almost every } x\in[0,1]\}.$$
Let $g\in M^\bot$, define $g_{even}(x):=\frac12(g(x)+g(-x))$ and $g_{odd}(x):=\frac12(g(x)-g(-x))$. Observe $g_{even}\in M$. Hence, we have $\int_{-1}^1 g(x)\cdot g_{even}(x)\ dx=0$. It follows
$$0= \int_{-1}^1g(x)g_{even}(x)\ dx= \int_{-1}^1(g_{even}(x))^2\ dx+\underbrace{\int_{-1}^1g_{even}(x)g_{odd}(x)\ dx}_{=0, \text{ shown in part a)}}= \int_{-1}^1(g_{even}(x))^2\ dx.$$
Hence, $g_{even}(x)=0$ for almost all $x\in[-1,1]$.
