Let $f:\Bbb{R^2}\to\Bbb{R}$ be a lebesgue measurable function, $f\in L^{\infty}$. For every $\theta\in [0,2\pi)$, define $g_\theta(x)=f(x\cos(\theta),x\sin(\theta))$. Prove that for almost every $\theta$, $g_\theta\in L^{\infty}$.
So, first I make a claim that $g_\theta$ is measurable - this is because it is the composition of a continuous function (which is Lebesgue measurable) and a Lebesgue measurable function and therefore Lebesgue measurable. I wasn't able to solve the second part. I thought maybe since $f$ is bounded a.e, then also $g_\theta$ must be bounded a.e, but I wasn't able to prove it.
Any hint would be appreciated.
Notice that the "polar coordinates" mapping $P: \mathbb{R}^2 \to (-\infty, \infty)\times [0,2\pi)$ is measurable, invertible with measurable inverse, and continuous almost everywhere. Moreover, let $\mu_2$ denote standard Lebesgue measure on $\mathbb{R}^2$, and $\mu_P$ denote standard Lebesgue measure on the product space $(-\infty, \infty)\times [0,2\pi)$. Then the measures $\mu_2$ and $\mu_P\circ P$ defined by $\mu_P\circ P(A) = \mu_P(P(A))$ are mutually absolutely continuous, that is $\mu_2(A) = 0$ if and only if $\mu_P \circ P(A) =0$ (why?).
The function $g:(-\infty, \infty)\times [0,2\pi) \to \mathbb{R}$ defined by $g(x,\theta) = f(xcos(\theta),xsin(\theta))=f(P^{-1}(x,\theta))$ is measurable, and further is an element of $L^\infty$, since for any $M > \|f\|_\infty$, $\mu_P( |g| > M) = \mu_P( |f \circ P^{-1}| > M ) = 0$, since $\mu_2(\{|f| > M \})=0$ (why?).
Now it follows by Fubini's theorem that the sections $g_\theta$ of $g$ must be in $L^\infty$ for almost all $\theta$. In particular by applying Fubini's theorem we have that $$0 = \mu_P( |g| > M) = \int_{(-\infty, \infty)\times [0,2\pi)} 1_{\{|g|>M\}}d\mu_P = \int_{[0,2\pi)}\left( \int_{\mathbb{R}} 1_{\{|g_\theta|>M\}} d\mu_1\right)d\mu([0,2\pi))$$, hence $\mu\{|g_\theta|>M\} =0$ for almost all $\theta \in [0,2\pi)$, which gives what we need.