What is Young measure?

Question

I read about Young measures from the book, Weak convergence methods for nonlinear partial differential equations by L.C. Evans. He introduces the concept by the following theorem:

Theorem. Assume that the sequence $\left\{f_{k}\right\}_{k=1}^{\infty}$ is bounded in $L^{\infty}\left(U ; \mathbb{R}^{m}\right) .$ Then there exists a subsequence $\left\{f_{k_{j}}\right\}_{j=1}^{\infty} \subset\left\{f_{k}\right\}_{k=1}^{\infty}$ and for a.e. $x\in U$ a Borel probability measure $\nu_{x}$ on $\mathbb{R}^{m}$ such that for each $F \in C\left(\mathbb{R}^{m}\right)$ we have $$ F\left(f_{k_{j}}\right) \stackrel{*}{\longrightarrow} \bar{F} \quad \text { in } \quad L^{\infty}(U), $$ where $$ \bar{F}(x) \equiv \int_{\mathbb{R}^{m}} F(y) d \nu_{x}(y) \quad(\text { a.e. } x \in U) $$ Definition. We call $\left\{\nu_{x}\right\}_{x \in U}$ the family of Young measures associated with the subsequence $\left\{f_{k_{j}}\right\}_{j=1}^{\infty}$.

My question is: what does $\nu_x(E)$ signify for any Borel set $E\subset \mathbb{R}^m$? Does it measure, for a fixed $x$, how often $f_k(x)$ takes value in $E$ as $k$ changes?

Answer 1

Related to Nate River's answer, I personally prefer to think of Young measures as single measures $\nu$ on $U \times \mathbb{R}^m$, which have the condition that their projection on the first component (the pushforward along $(x,y) \mapsto x$) is the Lebesgue measure on $U$. For any reasonable function $f:U\to \mathbb{R}^m$, you can define such a measure "living on" the graph of $f$ in $U \times \mathbb{R}^m$.¹

If you have a bounded sequence of functions $f_k$, using the usual compactness theorems for measures, there is a weakly converging subsequence of the corresponding measures (call them $\nu^{k_j}$) and because of the $L^\infty$ bounds, there is no mass escaping towards infinity, so the projection condition is conserved. The limit $\nu$ then is the Young measure limit (in the above sense) for that sequence.

In particular, now for $G \in C(U\times \mathbb{R}^m)$, we have $$\lim_{k_j \to \infty} \int_U G(x,f_{k_j}(x)) dx = \lim_{k_j \to \infty} \int_{U\times \mathbb{R}^m} G(x,y) d \nu^{k_j} = \int_{U\times \mathbb{R}^m} G(x,y) d \nu $$ just by definition and weak convergence. If you take $G(x,y) = F(y) \phi(x)$ for fixed $F$ and all $\phi$, you recover precisely the theorem in the question.

Now the "classic" Young measure $(\nu_x)_{x\in U}$ then consists of "vertical slices" of this measure, i.e. a disintegration. Since $\{x\} \times \mathbb{R}^m$ always has measure $0$ wrt. to any Young measure defined as above, you can also immediately see, that this is not well defined for any single $x$, but only if you consider enough of them.

If you think graphically about the measures $\nu^{k_j}$ and how they converge you can also see what the measure $\nu$ at $(x_0,y_0)$ (and thus the classic Young measure $\nu_{x_0}$ at $y_0$) represents, namely the limit of how often $f_{k_j}(x)$ for $x$ close to $x_0$ takes values $y$ close to $y_0$.²

There are many other interpretations of Young measures, but most of them are problem specific. If you know what $f_k$ represents, then this gives you a context of what the resulting Young measure limit is supposed to mean. (e.g. a classical example: If $f_k$ represents regularised solutions to a problem, where you force the oscillating stuff to stay on a scale $\frac{1}k$, then you can argue that $\nu_x$ represents the local microstructure, which is too small to resolve on the scale of functions)

There are also a lot of people that offer probabilistic interpretations of Young measures. Personally I am not a fan of those, because there is no randomness involved here. The only reason that $\nu_x$ is a probability measure is because that is the name we give to measures normalized to unit mass.

¹Roughly it is the $n$-dimensional Hausdorff measure restricted to the graph $\{(x,f(x)), x \in U\}$, locally scaled by a factor (something like $\sqrt{1+|\nabla f|^2}^{-1}$) to make the projection work. Alternatively, consider it the pushforward of the Lebesgue-measure on $U$ by $x\mapsto (x,f(x))$

²In particular the $x$ close to $x_0$ part is something that is often missed. Even if $f_{k_j}(x_0)$, converges to $y_0$, the Young measure $\nu_{x_0}$ can be completely different from $\delta_{y_0}$, because of what happens around it for different $x$. (Though only in a sense, as it is not strictly well defined in the first place).

Answer 2

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.