Conditions on continuity under Lebesgue measure

Question

Let $h : X \times I \rightarrow \mathbb{R}$ be a continuous function, where $X$ is a compact set of $\mathbb{R}^k$, for some $k$.

Set $\hat{h}(x,t) = 1$ if $h(x,t) \neq 0$, $0$ otherwise.

Define $g : I \rightarrow \mathbb{R}$ by $g(t) = \int_X \hat{h}(x,t) d\mu$, where $\mu$ is Lebesgue measure on $X$.

Under what conditions can we assert that $g$ is continuous?

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Leo, thank you for your effort in helping me find a solution. I think you missed part of the problem, though. The function $h$ definitely depends on both variables: $h(x,t)$. My question pertains to $\hat{h}$; not to $h$. $\hat{h}$ is defined in terms of $h$ as: $\hat{h}(x,t) = 1$ if $h(x,t) \neq 0$, $0$ otherwise.

If it helps clarity, fix $t$ at any value and set
$ A_t = \{x \in X \mid h(x, t) \neq 0\} $. Then an alternate definition of $g$ is, $g(t) = \mu(A_t)$. The question then is, "Is $g$ continuous as a function of $t$?"

Choose any $t_0 \in I$. It seems intuitive, since $h$ is continuous, that in some geometric sense we have

(I) $ A_t \rightarrow A_{t_0}$, and consequently that

(II) $\mu(A_t) \rightarrow \mu(A_{t_0})$,

which thus means that $g(t) \rightarrow g(t_0)$, in turn implying that $g$ is continuous.

The difficulty lies in capturing the geometric notion, (I), "measure-theoretically" so that we can assert (II).

Answer 1

We can equivalently define $\hat h$ as the indicator function of an open set $U\subset X\times I$, so $g$ is lower semi-continuous.

Set $A$ the boundary of $U$ and $V=(X\times I)\setminus U$. $U,A,V$ is a partition of $X\times I$, with $U,V$ open and $A$ closed. Set $g_V(t)=\int_X1_V(x,t)dx$, and $g_A=\int_X1_A(x,t)dx$. We have $g+g_A+g_V=1$, $g_U$ and $g_V$ are lower semi-continuous and $g_A$ is upper semi-continuous.

If $g_A$ is continuous, then $g+g_V=1-g_A$ is continuous, so $g$ and $g_V$ are continuous.

If $g_A$ is not continuous, then at least one of the other functions is discontinuous.

A case in which we can easily prove that $g_A$ is continuous is be the case $g_A=0$, that is "the trace of the boundary of $U$ on $X\times \{t\}$ is always negligible for the 1-dimensional Lesbesgue measure".