TITLE: If $b:[0,T]\times ℝ→ℝ$ has at most linear growth in $x$ and is Lipschitz continuous in $x$, then $t↦b(t,X(t))$ is continuous, if $X:ℝ→ℝ$ is continuous
QUESTION [1 upvotes]: Let

$T>0$
$b:[0,T]\times\mathbb R\to\mathbb R$ with $$|b(t,x)|^2\le K(1+|x|^2)\;\;\;\text{for all }t\in[0,T]\text{ and }x\in\mathbb R\tag 1$$ for some $K\ge 0$ and $$|b(t,x)-b(t,y)|\le L|x-y|\;\;\;\text{for all }t\in[0,T]\text{ and }x\in\mathbb R\tag 2$$ for some $L\ge 0$
$X:\mathbb R\to\mathbb R$ be continuous

Can we show that $$[0,T]\to\mathbb R\;,\;\;\;t\mapsto b(t,X(t))\tag 3$$ is continuous?
I've read that claim in a book, but I don't think that it's true. Clearly, the statement would be true, if $b$ wouldn't depend on the first variable. But otherwise, we should only be able to say that $$|b(s,X(s))-b(t,X(t))|\le 2L|X(s)-X(t)|+|b(s,X(t))-b(t,X(s))|\tag 4$$ for all $s,t\in[0,T]$; but I don't think that this inequality is helpful.

REPLY [1 votes]: This is false.  The function 
$$
b (t,x) = \begin {cases} 1 & t \in \Bbb {Q}, \\ 0, t \notin \Bbb {Q}\end {cases}
$$
fulfills your assumption for $K=1$ and $L=0$.
But if we take $X (t)\equiv 0$, we get a discontinuous function $t \mapsto b (t, X (t)) $.
The problem is that there is no assumption on the dependence of $b $ on $t $.
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