measure theory - A function constant almost everywhere

Q: Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is measurable with respect to Lebesgue measure and $f(x)=f(x+1)=f(x+\pi)$ for almost every $x$. Prove that $f$ is constant almost everywhere.

Proof attempt: Since the ratio of $1$ and $pi$ is irrational, given any real number $r$ we have there exist $m, n\in \mathbb{Z}$ such that $|r-(m+n \pi )|< \epsilon$, so that the set $P=\{m+n\pi : m, n\in \mathbb{Z}\}$ is everywhere dense.

Now by Lusin's theorem take the interval $[a, b]$, so there is a compact set $E\subset [a,b]$ such that $\mu(E) > (b-a-\delta)$ for some arbitrarily small $\delta$ and so that $f|_{E}$ is continous. Pick a point $x$ like the one in the condition of the problem and any $y\in E$. By continuity of $f$ on $E$ and by the density of $P$ there will be a point such that $|f(x)-f(y)|=|f(x)-f(y)|=|C-f(y)|<\epsilon_{2}$. Taking the limit as $\delta \rightarrow 0$, $f$ is constant almost everywhere on the interval $[a, b]$, and since the interval is arbitrary we can say $f$ is constant almost everywhere.

Does this look alright? Is there another way of doing this without invoking something like Lusin's theorem? Measurable functions can be very ugly, so are there any other results similar to Lusin's theorem that allow us to put some sort of regularity on measurable functions and gain some intuition about them?

Thanks in advance!

Answer

This answer explains my idea from the comment in more detail.

First, you can not truncate $f$ in such a way that the truncated version is in $L^1$, because this will destroy periodicity.

My idea was to consider

$$
g := f \cdot \chi_{|f| \leq n}.
$$

You should verify that $g$ is again "periodic" in the same sense as $f$ is (and bounded).

Now take any approximation of the identity (e.g. $h_\varepsilon = \varepsilon^{-n} \cdot h(x/\varepsilon)$, where $h \in C_c$ with $\int h \, dx = 1$) and consider the family of convoultion products

$$
F_\varepsilon := (g \ast h_\varepsilon) (x) = \int g(x-y) h_\varepsilon (y) \, dy.
$$

It is then not too hard to show that $g \ast h_\varepsilon$ is continuous and is periodic in the classical sense with periods $1,\pi$, i.e. $F_\varepsilon (x) = F_\varepsilon (x+1) = F_\varepsilon (x+\varepsilon)$ for all $x$. This is so, because the integral used in the definition of $F_\varepsilon$ does not "see" the null-set on which this property might fail for $g$ (or $f$).

Now conclude (using a variant of your proof) that $F_\varepsilon \equiv C_\varepsilon$ for some constant $C_\varepsilon$.

By classical theorems about approximation by convolution, you also get $F_\varepsilon (x) \to g(x)$ almost everywhere (at every Lebesgue-point of $g$, see e.g. Why convolution of integrable function $f$ with some sequence tends to $f$ a.e.).

This will allow you to conclude $g \equiv c$ almost everywhere and then also that $f \equiv c'$ almost everywhere for some constant $c'$.