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Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\int_{x\in\mathbb R} f(x)\ \mu(dx) \leq 1$ and that $g(x) \leq 1$ for all $x$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dx) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?

Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\int_{x\in\mathbb R} f(x)\ \mu(dx) \leq 1$ and that $g(x) \leq 1$ for all $x$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dx) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?

Edit:

If I switch the positions of $f$ and $g$ and only require uniformity w.r.t. $g$, then I obtain something that looks similar enough to a well-known result about $L_1$-convergence that someone, somewhere must have already studied it (though I haven't found any references to it yet). Accordingly, I've posted the simplified problem on math.stackexchange: L1-convergence in the Lebesgue differentiation theorem for general Radon measures? (I hope this was the right thing to do?)

Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\left\Vert f \right\Vert_1 \leq 1$ and $\left\Vert g \right\Vert_\infty \leq 1$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dx) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?

Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\int_{x\in\mathbb R} f(x)\ \mu(dx) \leq 1$ and that $g(x) \leq 1$ for all $x$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dx) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?

Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\left\Vert f \right\Vert_1 \leq 1$ and $\left\Vert g \right\Vert_\infty \leq 1$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dy) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?

Let $\mu$ be a finite measure on $\mathbb R$ and $f,g : \mathbb R \to \mathbb R_{\geq 0}$ two measurable maps such that $\left\Vert f \right\Vert_1 \leq 1$ and $\left\Vert g \right\Vert_\infty \leq 1$. If I understand Lebesgue's differentiation theorem correctly, then it is true that $$\lim_{\delta \to 0} \int_{x\in\mathbb R} f(x)\ \frac{\int_{y\in[x-\delta,x+\delta]} g(y) \ \mu(dy)}{\mu([x-\delta,x+\delta])} \mu(dx) = \int_{x\in\mathbb R} f(x)g(x) \ \mu(dx).$$

My initial question is the following: does anyone know if this convergence is uniform with respect to $f$ and $g$ (not $\mu$)? (i.e. do we have "$\forall\mu\ \forall \varepsilon\ \exists \delta\ \forall f\ \forall g\ \ldots$" instead of just "$\forall \mu\ \forall \varepsilon\ \forall f\ \forall g\ \exists \delta\ \ldots$"?)

I've taken a look at a couple of books that prove (generalisations of) Lebesgue's differentiation theorem and found no mention of something like that, so I understand that the answer is probably "no". What does give me some hope however are the results mentioned in this post: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, notably the "Lebesgue regularity lemma".

In the likely event that the answer to my initial question is "no", then does anyone know if uniformity can be obtained by adding constraints on $\mu$,$f$ and $g$? (e.g. requiring that they be "regular enough" in some sense?) Or do we at least have uniformity w.r.t $g$ but not $f$ or vice versa?