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Let $\mathcal{B}$ denote the Borel-$\sigma$-algebra with respect to the Euclidean topology $\mathcal{T}$. Show that the measure $\lambda_{(0,1)}$ is regular.

I start with outer regularity.

Let $B\in\mathcal{B}$. By the outer regularity of the Lebesgue-measure $\lambda$, we have $$ \lambda_{(0,1)}(B)=\lambda(B\cap (0,1))=\inf\left\{\lambda(U): B\cap (0,1)\subset U, U\in\mathcal{T}\right\} $$ Let $U\in\mathcal{T}$ such that $B\subset U$. Then $B\cap [0,1]\subset U$ and since $\lambda(U\cap [0,1])\leq\lambda (U)$, we should have $$ \inf\left\{\lambda(U): B\cap (0,1)\subset U, U\in\mathcal{T}\right\}\geq\inf\left\{\lambda(U\cap [0,1]): B\subset U, U\in\mathcal{T}\right\}\\ =\inf\left\{\lambda_{(0,1)}(U): B\subset U, U\in\mathcal{T}\right\}, $$ hence $$ \lambda_{(0,1)}(B)\geq \inf\left\{\lambda_{(0,1)}(U): B\subset U, U\in\mathcal{T}\right\} $$

But how to show "$\leq$"? And inner regularity?

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