Does $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ hold for Lebesgue integrals? This question is related to this post. Proving that $\int_{0}^{a} f(x) \;\mathrm dx = \int_{0}^{a} f(a - x) \;\mathrm dx$
It is proven in the answers to the post (at least for Riemann integrals) that $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. I was wondering whether this also holds for Lebesgue integrals. I suppose the corresponding statement could be written as
$$\int_{[0,a]} f(x) d\mu(x) = \int_{[0,a]} f(a-x) d\mu(x),$$
where $\mu$ is the Lebesgue measure.
I am guessing that if this holds true, it can be proven first by simple functions and then by extending the proof in the usual way. It has been quite some time since I studied Lebesgue integrals, and when I tried to show this by simple functions I ended up confusing myself.
Should this statement be true, I would very much appreciate a proof or a proof sketch. Should it not be true, I would also very much appreciate a counterexample. Thanks in advance to all that took the time to read this post. Apologies if the question is not clear.
Edit:
My attempt has been something like this. Consider $f$ as a simple function and write
$f(x) = a_{1}$ for $x\in E_{1}$, ..., $f(x) = a_{k}$ for $x\in E_{k}$, where $E = [0, a] = \cup_{k} E_{k}$. Then
$$\int_{[0, a]} f d\mu = \sum_{j=1}^{k} a_{j}\mu(E_{j}).$$
Similarly we could write $f(a-x) = a_{1}$ for $a-x \in E_{1}$, ..., $f(a-x) = a_{k}$ for $a-x \in E_{k}$. I tried to show equality using this setup, but I kept messing up.
Edit 2:
Could one then write that $f(a-x) = a_{1}$ for $-x \in E_{1} - a$, ... , $f(a-x) = a_{k}$ for $-x \in E_{k} - a$, which then gives us
$$\int_{E} f(a-x)d\mu(x) = \sum_{j=1}^{k} a_{j} \mu(E_{j} - a) = \sum_{j=1}^{k} a_{j} \mu(E_{j}) = \int_{E} f(x)d\mu(x) ?$$
Something feels a bit off with this attempt.
 A: The key fact is that if $E\subset[0,a]$, then $a-x\in E$ if and only if $x\in a-E := \{a-y\mid y\in E\}$. If $E\subset [0,a]$ is measurable, then so is $a-E$, and by the key fact,
$$
\int_{[0,a]}1_E(a-x)\,d\mu(x) = \int_{[0,a]}1_{a-E}(x)\,d\mu(x) = \mu((a-E)\cap [0,a]).
$$
Of course, since $E\subset [0,a]$, $a-E\subset [0,a]$, so $\mu((a-E)\cap [0,a]) = \mu(a-E)$. By translation and reflection invariance of the Lebesgue measure, $\mu(a-E) = \mu(E) = \int_0^a1_E(x)\,d\mu(x)$.
If $\phi$ is a simple function supported in $[0,a]$, then linearity implies $$\int_{[0,a]}\phi(a-x)\,d\mu(x) = \int_{[0,a]}\phi(x)\,d\mu(x).$$ For nonnegative measurable functions $f$ supported in $[0,a]$, the result follows by monotone convergence applied to a sequence of simple functions $0\le \phi_n\nearrow f$ pointwise, and for general integrable functions $f$ supported in $[0,a]$, apply the result obtained for nonnegative functions to the canonical decomposition $f = f^+- f^-$, and use linearity once more. (By definition, $f^+ = \max(f,0)$, and $f^- = \min(f,0)$.)
