If $u \in L^\infty((0,T)\times \Omega)$, does it follow that $u(t) \in L^\infty(\Omega)$ for a.e $t$? Let $u \in L^\infty((0,T)\times \Omega)$, where $\Omega \subset \mathbb{R}^n$ is a smooth bounded domain. We know that it is not necessarily true that $u$ belongs to $L^\infty(0,T;L^\infty(\Omega))$. 
But does it still follow that $u(t) \in L^\infty(\Omega)$ for a.e. $t \in (0,T)$?
By Fubini theorem, I thought $\int_\Omega |u(t,x)|^p\;\mathrm{d}x$ should exist for a.e. $t$ and almost every $p$. But I don't see that this helps me
 A: For simplicity, let us assume that the measure of $\Omega$ is $1$.
For an arbitrary, measurable subset $A \subset (0,T)$, we have
$$
(\int_A \| u(t)\|_{L^p(\Omega)} \,\mathrm{d}t)^p \, \mu(A)^{-p/q}
\le
\int_A \| u(t)\|_{L^p(\Omega)}^p \,\mathrm{d}t
=
\int_A \int_\Omega |u|^p
=
\int_{A \times \Omega} |u|^p
\le
\|u\|_{L^\infty((0,T)\times\Omega)}^p.
$$
Here, we used Hölder and Fubini and $1/p + 1/q = 1$.
Hence
$$\int_A \|u(t)\|_{L^p(\Omega)} \le \mu(A)^{1/q} \, \|u\|_{L^\infty((0,T) \times \Omega)}$$
Note that the left-hand side integrand is monotone w.r.t. $p$.
With $p \to \infty$ and $q \to 1$ we get
$$\int_A \|u(t)\|_{L^\infty(\Omega)} \le \mu(A) \, \|u\|_{L^\infty((0,T) \times \Omega)}.$$
This implies that the function
$t \mapsto \|u(t)\|_{L^\infty(\Omega)}$ is essentially bounded by $\|u\|_{L^\infty((0,T) \times \Omega)}$.
A: Let me rephrase the question and state it in a more general form.
Proposition:
Let $(X,\mathcal{M}_{X},\mu_{X})$, $(Y,\mathcal{M}_{Y},\mu_{Y})$
be measure spaces, where $\mu_{X}$ and $\mu_{Y}$ are $\sigma$-finite
measures. Let $(X\times Y,\mathcal{M}_{X}\otimes\mathcal{M}_{Y},\mu_{X}\times\mu_{Y})$
be the product measure space. Then for each $f\in\mathcal{\mathcal{L}^{\infty}}(X\times Y,\mathcal{M}_{X}\otimes\mathcal{M}_{Y},\mu_{X}\times\mu_{Y})$,
there exists $B\in\mathcal{M}_{X}$ with $\mu_{X}(B^{c})=0$ such
that for each $x\in B$, $f(x,\cdot)\in\mathcal{L}^{\infty}(Y,\mathcal{M}_{Y},\mu_{Y})$.
///////////////////////////////////////////
Proof: Let $f\in\mathcal{\mathcal{L}^{\infty}}(X\times Y,\mathcal{M}_{X}\otimes\mathcal{M}_{Y},\mu_{X}\times\mu_{Y})$.
Claim 1: Let $x_{0}\in X$ be arbitrary. We assert that $f(x_{0},\cdot)$
is $\mathcal{M}_{Y}/\mathcal{B}(\mathbb{R})$-measurable. (This part is not essential and
may be skipped) Let $\pi_{X}:X\times Y\rightarrow X$ and $\pi_{Y}:X\times Y\rightarrow Y$
be the canonical projections. Then $Y\rightarrow X\times Y$, $y\mapsto(x_{0},y)$
is $\mathcal{M}_{Y}/\mathcal{M}_{X}\otimes\mathcal{M}_{Y}$-measurable.
For, we denote this map by $\theta$ temporarily. Observe that $\pi_{X}\circ\theta:Y\rightarrow X$
is the constant map $y\mapsto x_{0}$, which is $\mathcal{M}_{X}/\mathcal{M}_{Y}$-measurable,
and $\pi_{Y}\circ\theta:Y\rightarrow Y$ is the identity map $y\mapsto y$,
which is $\mathcal{M}_{Y}/\mathcal{M}_{Y}$-measurable. Since both
$\pi_{X}\circ\theta$ and $\pi_{Y}\circ\theta$ are measurable, by
the universal property of product $\sigma$-algebra, it follows that
$\theta$ is $\mathcal{M}_{Y}/\mathcal{M}_{X}\otimes\mathcal{M}_{Y}$-measurable.
Finally, $f(x_{0},\cdot)=f\circ\theta$ and hence $f$ is $\mathcal{M}_{Y}/\mathcal{B}(\mathbb{R})$-measurable.
/////////////////////////////////////////
Core part: By definition, $f$ is essentially bounded. That is, there
exists $M>0$ and $A\in\mathcal{M}_{X}\otimes\mathcal{M}_{Y}$ with
$\mu_{X}\times\mu_{Y}(A)=0$ such that $|f(x,y)|\leq M$ for all $(x,y)\in A^{c}$.
Define $g:X\rightarrow[0,\infty]$ and $h:Y\rightarrow[0,\infty]$
by 
\begin{eqnarray*}
g(x) & = & \int1_{A}(x,y)d\mu_{Y}(y),\\
h(y) & = & \int1_{A}(x,y)d\mu_{X}(x).
\end{eqnarray*}
By Tonelli-Fubini Theorem, $g$ is $\mathcal{M}_{X}/\mathcal{B}$-measurable,
$h$ is $\mathcal{M}_{Y}/\mathcal{B}$-measurable. Moreover $\int g(x)d\mu_{X}(x)=\int h(y)d\mu_{Y}(y)=\mu_{X}\times\mu_{Y}(A)=0$.
Since $\int g(x)d\mu_{X}(x)=0$, there exists $B\in\mathcal{M}_{X}$
with $\mu_{X}(B^{c})=0$ such that $g(x)=0$ for all $x\in B$. For
each $x\in B$, $g(x)=0\Rightarrow\mu_{Y}\left(A_{x}\right)=0$, where
$A_{x}=\{y\in Y\mid(x,y)\in A\}$, because $1_{A_{x}}(y)=1_{A}(x,y)$.
(This also shows that $A_x \in \mathcal{M}_Y$ by Claim 1.)
Finally, for each $x\in B$, $f(x,\cdot)$ is essentially bounded
because $\mu_{Y}(A_{x})=0$ and for each $y\in(A_{x})^{c}$, we have
$(x,y)\in A^{c}\Rightarrow|f(x,y)|\leq M$.
