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I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2-x_1|, |y_2^{1/3}-y_1^{1/3}|) $$ (if $y<0$, then $y^{1/3}$ just means $-|y|^{1/3}$).

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2-x_1|, |y_2^{1/3}-y_1^{1/3}|) $$ (if $y<0$, then $y^{1/3}$ just means $-|y|^{1/3}$).

Update: This answer was originally based on my recollection of the "standard fact" that the higher-dimensional Lebesgue differentiation theorem fails for rectangles if the eccentricity is not restricted. This much is true if arbitrary rectangles are allowed, but of course the situation here is different, and I'm not sure now what the situation is (and in fact I'm not even sure it's not an open question).

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max ((x_2-x_1)^2, |y_2-y_1|) . $$

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2-x_1|, |y_2^{1/3}-y_1^{1/3}|) $$ (if $y<0$, then $y^{1/3}$ just means $-|y|^{1/3}$).

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2^2-x_1^2|, |y_2-y_1|) . $$

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max ((x_2-x_1)^2, |y_2-y_1|) . $$