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This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic - consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaogu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost... up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) := \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i} (E)$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”:

$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $

with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ - replace the test sets $E$ with continuous functions vanishing at infinity and use a seperability/diagonalization argument to properly define the limit measures.

However I am not sure if you can take limits as $|A_i| \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though...

Still, I hope this has somewhat helped at least on a heuristic level.

This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic — consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaoglu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost… up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) \mathrel{:=} \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i} (E)$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”: $$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $$ with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ — replace the test sets $E$ with continuous functions vanishing at infinity and use a seperability/diagonalization argument to properly define the limit measures.

However I am not sure if you can take limits as $\lvert A_i\rvert \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though….

Still, I hope this has somewhat helped at least on a heuristic level.

This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic - consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaogu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost... up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) := \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i}$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”:

$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $

with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ - replace the test sets $E$ with continuous functions vanishing at infinity and use a seperability/diagonalization argument to properly define the limit measures.

However I am not sure if you can take limits as $|A_i| \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though...

Still, I hope this has somewhat helped at least on a heuristic level.

This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic - consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaogu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost... up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) := \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i} (E)$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”:

$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $

with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ - replace the test sets $E$ with continuous functions vanishing at infinity and use a seperability/diagonalization argument to properly define the limit measures.

However I am not sure if you can take limits as $|A_i| \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though...

Still, I hope this has somewhat helped at least on a heuristic level.

This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic - consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaogu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost... up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) := \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i}$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”:

$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $

with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ - replace the test sets $E$ with continuous compactly supported functions and use a seperability/diagonalization argument to properly define the limits.

However I am not sure if you can take limits as $|A_i| \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though...

Still, I hope this has somewhat helped at least on a heuristic level.

This is probably more fitting for a comment, but it’s a little long.

Maybe this might be a helpful heuristic - consider, instead of points $x \in U$, some countable partition of $U$ into sets $A_i$ of non null measure. We will define the “discrete Young measures” $\nu_{A_i}$.

Now to each $f_{k}$ is associated a measure $\mu_{k}$ on $\mathbb R^m$ via the pullback. Banach Alaogu gives us a weakly-* converging subsequence $\mu_{k_j}$. Cheating a little we can say that for some $\mu$, $\mu_{k_j}(E) \to \mu(E)$, almost... up to wobbling $E$ a little.

Now $\mu_{k_j}(E)$ by definition is $\mathcal L (f_{k_j}^{-1}(E))$, where $\mathcal L$ is the Lebesgue measure on $U$. Since the $A_i$ form a partition of $U$, we have $\mathcal L (f_{k_j}^{-1}(E)) = \sum_i \mathcal L (f_{k_j}^{-1}(E) \cap A_i) := \sum_i \mu^i_{k_j}(E)$.

Heuristically, this means that each $A_i$ contributes a certain amount to the mass $\mu_{k_j}(E)$ according to how much time $f_{k_j}$ (restricted to $A_i$) spends in $E$.

Since $\mu_{k_j}(E)$ “converges” to $\mu(E)$ we have that $\mu_{k_j}^i(E)$ converges pointwise in $i$ to some $\hat \mu^i (E)$. What this means is that asymptotically, the $f_{k_j}$ restricted to $A_i$ spend a constant amount of time in $E$. We take $\nu_{A_i}$ to be $\hat \mu^i (E)$.

So the “Young measure” $\nu_{A_i}$ has the effect of smoothing out oscillations in the $f_{k_j}$. Even though $f_{k_j}$ can vary wildly within $A_i$, the distribution of values of $f_{k_j}$ restricted to $A_i$ converges to $\nu_{A_i}$.

Note also we have the “integration formula”:

$ \mu(E) = \int_{\mathbb N} \int_{\mathbb R^m} 1_E (y) \ d \nu_{A_i}(y) \ d\mathcal C(i) $

with $\mathcal C$ being the counting measure. This is the discrete counterpart of the formula for $\bar F$ in your post.

The above can be carried out more rigorously to actually properly define the discrete Young measures $\nu_{A_i}$ - replace the test sets $E$ with continuous functions vanishing at infinity and use a seperability/diagonalization argument to properly define the limit measures.

However I am not sure if you can take limits as $|A_i| \to 0$ to define the proper Young measures $\nu_x$, which is why the actual construction uses the machinery of measure disintegration. It may be possible though...

Still, I hope this has somewhat helped at least on a heuristic level.