# Continuity of Lebesgue integral with continuously-varying measures?

Consider the locally bounded mapping $m: X \times \mathcal{B}(X) \rightarrow [0,1]$, with $X \subseteq \mathbb{R}^n$ and $\mathcal{B}(X)$ denoting the Borel sets, such that

• for all $x \in X$, $\ m(x,\cdot)$ is a probability measure on $X$, so that $\forall x \in X$ $\ m(x,X)=1$;

• for all $\tilde{X} \subseteq X$, the mapping $m(\cdot,\tilde{X})$ is continuous.

Given a continuous function $\ f: X \rightarrow [0,1]$, I am wondering on the following integral to be a continuous function as well.

$$x \mapsto F(x) := \int_X f(y) m(dy,x)$$

Comment. This is a variation of that question, where for "constant" probability measure $m$, the Dominated Converge Theorem is sufficient to show continuity of the integral. Now I think it is interesting to ask if this holds whenever the probability measure $m$ changes continuously. I believe continuity does not hold, but I was not able to find a counterexample.

Edit. $f$ takes values on $[0,1]$.

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Thanks for your comments. I forgot to say that $f$ takes values on $[0,1]$. But I ignore what Baire class one is. – Adam Apr 14 '13 at 10:14

When $f$ is bounded, we can approximate it uniformly by simple functions, that is, linear combinations of characteristic functions of measurable sets, that is, of the form $\sum_{j=1}^Nc_j\chi_{A_j}$ where $c_j$ are constant and $\chi_{A_j}(x)$ is $1$ when $x\in A_j$ and $0$ otherwise. This only require $f$ to be bounded, no matter what the domain is.

For such functions, the associated $F$ is continuous by assumption (it is true when $f$ is the characteristic function of a measurable set by the second bullet, and a linear combination of continuous functions is continuous). Call $(f_n,n\geqslant 1)$ the sequence of simple functions converging uniformly to $f$ on $X$, and $(F_n,n\geqslant 1)$ the associated sequence like in the OP.

The sequence $(F_n,n\geqslant 1)$ uniformly converges to $F$ by the assumption that $m(x,\cdot)$ is a probability measure. Indeed, we have $$|F_n(x)-F(x)|\leqslant \int_X|f_n(y)-f(y)|m(dy,x)\leqslant \sup_{y\in X}|f_n(y)-f(y)|\cdot \underbrace{m(X,x)}_{=1}$$

Define for any measurable $g$, integrable with respect to all $m(\cdot,x)$, $$T(g)(x):=\int_Xg(y)m(dy,x).$$ We have that $T(f_n)$ is continuous for all $n$. Indeed, since $T$ is linear and $f_n=\sum_{\mbox{finite}}c_j\chi_{A_j}$, it's enough to prove that for any $S\subset X$ measurable, $T(\chi_S)$ is continuous. We have $T(\chi_S)(x)=m(S,x)$ which gives a continuous map by the second bullet in the OP.

Since an uniform limit of continuous functions is continuous, so will be $F$.

When $f$ is not bounded, by a truncation argument we can see that $F$ is Baire class one, that is, a pointwise limit of continuous functions.

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Thanks for your answer. Can you please be more specific in the first part? What do you mean by characteristic functions? Is compactness of $X$ required to "approximate" $f$ uniformly? What do you mean by "uniform limit of continuous functions"? – Adam Apr 14 '13 at 10:23
Can you please explain why "For such functions, the associated F is continuous by assumption"? I mean if $f$ is some indicator function, "from which assumption" we have that the corresponding $F$ is continuous as well? We have to prove this, right? – Adam Apr 14 '13 at 10:34
It's from the second bullet in your question. – Davide Giraudo Apr 14 '13 at 10:39
Sure. But why this assumption implies that $F$ is continuous? Moreover, we cannot take indicator functions in place of $f$, because $f$ is continuous. – Adam Apr 14 '13 at 10:41
That's why we approximate by simple function. For simple functions, we can see it's true, and uniform limit preserves continuity. – Davide Giraudo Apr 14 '13 at 10:42