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If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is Borel. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction.

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is Borel. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous.

One can also come up with examples where $f$ is $\mu$-measurable and discontinuous, and every linear functional which is $\mu$-a.e. equal to $f$ is also discontinuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction. (Of course, as a consequence of measurability, there will be a Borel function $g$ with $f=g$ a.e.; but then $g$ will not be linear!)

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is closed. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction.

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is Borel. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction.

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is closed. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense Borel subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction.

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is closed. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction.