The aim is to show that $$ \int_A f(x) \, dx \leq \int_A f(x +a) \, dx $$
holds for any Borel-set $A$, because this is equivalent to $f(x) \leq f(x +a)$ almost surly. We will show that this holds first for open intervals, then for any open set, and then for any borel-set.
Its enough to prove the inequality for $f^+$ and $f^-$ seperatly, hence it is sufficent to only prove the case where $f \geq 0$.
Step 1: Open Intervals
Take $a \in \mathbb{R}_+$. Then there exists a sequence $(a_n)_{n \in \mathbb{N}} \subset \mathbb{Q}$ with $a_n \uparrow a$ with
$$ f(x+ a_n) \geq f(x) .$$
In particular, we get $$ \int_I f(x) \, dx \leq \int_I f(x +a_n) \, dx $$ for each interval $I=[x_1,x_2]$. Further, since $f \in L_{loc}^{1}(\mathbf{R})$ we get from the dominated convergence theorem
$$ \int_I f(x+ a_n) \, dx = \int_{x_1}^{x_2} f(x + a_n) \, dx = \int_{x_1 + a_n}^{x_2 + a_n} f(x ) \, dx \to \int_{x_1 + a}^{x_2 + a} f(x) \, dx = \int_I f(x+ a) dx. $$
Thus we have
$$\int_I f(x) \, dx \leq \int_I f(x+a) \, dx $$ for each interval $[x_1,x_2]$. Since $\lambda(\{x_1,x_2\})=0$ we have $$\int_{[x_1,x_2]} f(x) \, dx = \int_{
(x_1,x_2)} f(x) \, dx $$ and thus the above inequality is also true for any open interval.
Step 2: Open Sets
Given any open set $A$. We find a countable union of disjoint open intervals $I_i$ with $$ A = \bigcup_{i \in \mathbb{N}} I_i $$
(Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]).
Since $f \geq 0$ we get
$$ \int_A f(x) \, dx = \sum_{i \in \mathbb{N}} \int_{I_i} f(x) \, dx \leq \sum_{i \in \mathbb{N}} \int_{I_i} f(x+a_n) \leq \int_A f(x+a_n) $$
from the monotoic convergence theorem. Note that both sides of the equation might be $\infty$.
Step 3: Borel-sets
Given any Borel-set $A$. From the outer regularity you can find a sequence of open sets $U_i$ with $1_{U_i} \downarrow 1_A$. Using the monotonic convergence gives the expected result.
