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Let $\sigma: \mathbb R^d \to \mathbb R^{d \times d}$ be of bounded variation.

Suppose $\sigma$ is uniformly elliptic, in the sense that there exists some constant $C > 0$ such that $\xi^{T} \sigma(x) \sigma(x)^{T} \xi \geq C |\xi|^2$ for all $x, \xi \in \mathbb R^d$.

It is known (see eg, Evans and Gariepy, Measure Theory and Fine Properties of Functions, Theorem 6.12) that since $\sigma$ is of bounded variation, there exists a sequence $\sigma_n$ of Lipschitz continuous functions such that $\mu(\{x| \sigma_n (x) \neq \sigma(x)\})\to 0$, where $\mu$ denotes the Lebesgue measure.

Question: Can $\sigma_n$ always be taken to be uniformly elliptic, with a possibly smaller ellipticity constant? I require that the constant hold for all $\sigma_n$ at once.

Let $\sigma: \mathbb R^d \to \mathbb R^{d \times d}$ be of bounded variation.

Suppose $\sigma$ is uniformly elliptic, in the sense that there exists some constant $C > 0$ such that $\xi^{T} \sigma(x) \sigma(x)^{T} \xi \geq C |\xi|^2$ for all $x, \xi \in \mathbb R^d$.

It is known (see eg, Evans and Gariepy, Measure Theory and Fine Properties of Functions, Theorem 6.12) that since $\sigma$ is of bounded variation, there exists a sequence $\sigma_n$ of (locally) Lipschitz continuous functions such that $\mu(\{x| \sigma_n (x) \neq \sigma(x)\})\to 0$, where $\mu$ denotes the Lebesgue measure.

Question: Can $\sigma_n$ always be taken to be uniformly elliptic, with a possibly smaller ellipticity constant? I require that the constant hold for all $\sigma_n$ at once.

Let $\sigma: \mathbb R^d \to \mathbb R^{d \times d}$ be of bounded variation.

Suppose $\sigma$ is uniformly elliptic, in the sense that there exists some constant $C > 0$ such that $\xi^{T} \sigma(x) \sigma(x)^{T} \xi \geq C |\xi|^2$ for all $x, \xi \in \mathbb R^d$.

It is known (see eg, Evans and Gariepy, Measure Theory and Fine Properties of Functions, Theorem 6.12) that since $\sigma$ is of bounded variation, there exists a sequence $\sigma_n$ of Lipschitz continuous functions such that $\mu(\{x| \sigma_n (x) \neq \sigma(x)\})\to 0$ where $\mu$ denotes the Lebesgue measure.

Question: Can $\sigma_n$ always be taken to be uniformly elliptic, with a possibly smaller ellipticity constant? I require that the constant hold for all $\sigma_n$ at once.

Let $\sigma: \mathbb R^d \to \mathbb R^{d \times d}$ be of bounded variation.

Suppose $\sigma$ is uniformly elliptic, in the sense that there exists some constant $C > 0$ such that $\xi^{T} \sigma(x) \sigma(x)^{T} \xi \geq C |\xi|^2$ for all $x, \xi \in \mathbb R^d$.

It is known (see eg, Evans and Gariepy, Measure Theory and Fine Properties of Functions, Theorem 6.12) that since $\sigma$ is of bounded variation, there exists a sequence $\sigma_n$ of Lipschitz continuous functions such that $\mu(\{x| \sigma_n (x) \neq \sigma(x)\})\to 0$, where $\mu$ denotes the Lebesgue measure.

Question: Can $\sigma_n$ always be taken to be uniformly elliptic, with a possibly smaller ellipticity constant? I require that the constant hold for all $\sigma_n$ at once.