[proofplan] The forward direction ($\Rightarrow$) reduces measurability plus truncation to the Lebesgue point theorem. We truncate $f$ to bounded functions $f_k = \max(-k, \min(f, k))$, each of which lies in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$ and hence is approximately continuous $\mathcal{L}^n$-a.e. by [Lebesgue Points Are Approximately Continuous](/theorems/3033). At points where $|f(x)| < k$, the function $f$ agrees with $f_k$ locally, so approximate continuity of $f_k$ transfers to $f$. The backward direction ($\Leftarrow$) uses Lusin's theorem in reverse: approximate continuity at a point implies that $f$ is the pointwise limit of continuous functions along a density-$1$ set, and we construct the approximating sequence using the approximate limit definition, showing that the preimage $f^{-1}((a, \infty))$ is measurable for every $a \in \mathbb{R}$. [/proofplan]

[step:Forward direction: truncate $f$ to produce bounded locally integrable functions] Assume $f : \mathbb{R}^n \to \mathbb{R}$ is $\mathcal{L}^n$-measurable. For each $k \in \mathbb{N}$, define the truncation \begin{align*} f_k : \mathbb{R}^n &\to [-k, k] \\ x &\mapsto \max(-k, \min(f(x), k)). \end{align*} Each $f_k$ is measurable (as a composition of measurable and continuous functions) and bounded by $k$. For any compact set $K \subset \mathbb{R}^n$, we have $\int_K |f_k| \, d\mathcal{L}^n \leq k \cdot \mathcal{L}^n(K) < \infty$, so $f_k \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. [/step]

[step:Apply the Lebesgue point theorem to each $f_k$ and transfer approximate continuity to $f$] By the theorem [Lebesgue Points Are Approximately Continuous](/theorems/3033) applied to $f_k \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$, the function $f_k$ is approximately continuous at $\mathcal{L}^n$-almost every point. Let $N_k$ denote the $\mathcal{L}^n$-null set where $f_k$ fails to be approximately continuous. Define $A_k := \{x \in \mathbb{R}^n : |f(x)| < k\}$. On $A_k$, we have $f(x) = f_k(x)$ (since $|f(x)| < k$ means the truncation does not activate). Moreover, for $x \in A_k$, the set $\{y : |f(y)| < k\}$ contains $x$ and is measurable. By the [Lebesgue Density Theorem](/theorems/3023), $\mathcal{L}^n$-almost every $x \in A_k$ is a density point of $A_k$, meaning \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{L}^n(A_k \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = 1. \end{align*} Let $D_k$ denote the $\mathcal{L}^n$-null set of points of $A_k$ that are not density points of $A_k$. For $x \in A_k \setminus (N_k \cup D_k)$, we claim $f$ is approximately continuous at $x$. Fix $\varepsilon > 0$. Since $f_k$ is approximately continuous at $x$ with value $f_k(x) = f(x)$: \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{L}^n(\{y \in B(x,r) : |f_k(y) - f(x)| \geq \varepsilon\})}{\mathcal{L}^n(B(x,r))} = 0. \end{align*} On the set $A_k \cap B(x,r)$, we have $f = f_k$, so $\{y \in A_k \cap B(x,r) : |f(y) - f(x)| \geq \varepsilon\} \subset \{y \in B(x,r) : |f_k(y) - f(x)| \geq \varepsilon\}$. Therefore \begin{align*} \{y \in B(x,r) : |f(y) - f(x)| \geq \varepsilon\} &\subset \{y \in B(x,r) : |f_k(y) - f(x)| \geq \varepsilon\} \cup (B(x,r) \setminus A_k). \end{align*} Dividing by $\mathcal{L}^n(B(x,r))$ and taking $r \to 0^+$: the first term tends to $0$ by approximate continuity of $f_k$, and \begin{align*} \frac{\mathcal{L}^n(B(x,r) \setminus A_k)}{\mathcal{L}^n(B(x,r))} = 1 - \frac{\mathcal{L}^n(A_k \cap B(x,r))}{\mathcal{L}^n(B(x,r))} \to 1 - 1 = 0 \end{align*} since $x$ is a density point of $A_k$. Hence the density of $\{y : |f(y) - f(x)| \geq \varepsilon\}$ at $x$ is $0$, so $f$ is approximately continuous at $x$. [/step]

[step:Conclude the forward direction] Define $N = \bigcup_{k=1}^\infty (N_k \cup D_k)$, a countable union of $\mathcal{L}^n$-null sets, so $\mathcal{L}^n(N) = 0$. For any $x \in \mathbb{R}^n \setminus N$, since $f(x)$ is finite, there exists $k \in \mathbb{N}$ with $|f(x)| < k$. Then $x \in A_k \setminus (N_k \cup D_k)$, and the previous step shows $f$ is approximately continuous at $x$. Therefore $f$ is approximately continuous at $\mathcal{L}^n$-almost every point. [/step]

[step:Backward direction: show approximate continuity a.e. implies measurability] Assume $f : \mathbb{R}^n \to \mathbb{R}$ is approximately continuous at $\mathcal{L}^n$-almost every point. Let $N$ be the $\mathcal{L}^n$-null set where approximate continuity fails, and let $G = \mathbb{R}^n \setminus N$. We show $f^{-1}((a, \infty))$ is $\mathcal{L}^n$-measurable for every $a \in \mathbb{R}$, which suffices for measurability. By [Lusin's Theorem](/theorems/3015) applied to the restriction $f|_G$ (which can be shown to be measurable by the argument below), it suffices to verify that the preimages under $f$ of open sets are measurable. We proceed directly. Fix $a \in \mathbb{R}$ and let $E = \{x \in G : f(x) > a\}$. We claim $E$ is measurable. For each $x \in E$, set $\varepsilon = f(x) - a > 0$. Since $f$ is approximately continuous at $x$, the set $\{y : |f(y) - f(x)| < \varepsilon\} \subset \{y : f(y) > a\}$ has density $1$ at $x$. In particular, $x$ is a density point of the set $\{y : f(y) > a\}$. For each $m \in \mathbb{N}$, define \begin{align*} E_m := \{x \in G : f(x) > a + 1/m\}. \end{align*} Then $E = \bigcup_{m=1}^\infty E_m$. We show each $E_m$ is measurable. For $x \in E_m$, approximate continuity at $x$ with $\varepsilon = 1/(2m)$ gives that $\{y : f(y) > a + 1/(2m)\}$ has density $1$ at $x$. Define \begin{align*} F_m := \{x \in \mathbb{R}^n : \text{the set } \{y : f(y) > a + 1/(2m)\} \text{ has density } 1 \text{ at } x\}. \end{align*} The set $F_m$ is $\mathcal{L}^n$-measurable (it is defined by a limsup condition on measurable sets). We have $E_m \subset F_m$ by the argument above. Conversely, if $x \in F_m$ and $x \in G$, then the approximate limit of $f$ at $x$ equals $f(x)$, and the density-$1$ condition forces $f(x) \geq a + 1/(2m) > a$. Therefore $E_m = F_m \cap G$ up to a subset of $N$ (which is null). Since $F_m$ and $G$ are measurable, $E_m$ is measurable. Hence $E = \bigcup_m E_m$ is measurable, and $f^{-1}((a, \infty)) = E \cup (N \cap f^{-1}((a,\infty)))$. Since $N$ is a null set (hence measurable) and subsets of null sets are measurable (by completeness of Lebesgue measure), $f^{-1}((a, \infty))$ is $\mathcal{L}^n$-measurable. [/step]