Young measure

Remarks

A sequence of measurable $u_k \colon \Omega \to \R^m$ generates $(\nu_x)$ if $\varphi(u_k) \to \int \varphi \, d\nu_x$ weakly-$*$ in $L^\infty(\Omega)$ for every $\varphi \in C_0(\R^m)$. The fundamental theorem on Young measures says that every sequence bounded in $L^\infty(\Omega; \R^m)$ has a subsequence that generates a Young measure. Example: a measurable $u \colon \Omega \to \R^m$ gives the Young measure $\nu_x = \delta_{u(x)}$, since $\int \varphi \, d\nu_x = \varphi(u(x))$ is measurable. Example: $u_k(x) = \sgn(\sin(2 \pi k x))$ on $(0,1)$ generates the constant family $\nu_x = \frac12 (\delta_{-1} + \delta_1)$. Non-example: if $g \colon \Omega \to \R^m$ is not measurable, then $\nu_x = \delta_{g(x)}$ is not a Young measure. Some open $U$ has a non-measurable preimage, and $1_U$ is an increasing limit of functions $\varphi_j \in C_0(\R^m)$, so some $\varphi_j \circ g$ is not measurable.