Integral of a Function Defined on a Point

Question

Let $(\mathcal{X}, F_{\mathcal{X}}, \nu)$ be a $\sigma$-finite measure space and fix $y_0 \in \mathbb{R}^n$. We suppose that $g : \mathcal{X} \times \mathbb{R}^n \to \mathbb{R}$ is a product space measurable function such that $\int_{\mathcal{X}}g(x,y)d\nu(x)$ is defined and $C^{\infty}$ function of $y$ in a neighborhood of $y_0$ .

Now let $f:\mathcal{X} \to \mathbb{R}$ be a measurable function such that :

\begin{equation*} \int_{\mathcal{X}} |f(x)|g(x,y_0)d\nu(x) < \infty \end{equation*}

Does it follow that, for all $y$ in a neighborhood of $y_0$, we have :

\begin{equation*} \int_{\mathcal{X}} |f(x)|g(x,y)d\nu(x) < \infty \end{equation*}

If the results is not positive, what if we assume furthermore for every $x \in \mathcal{X}$ that $g(x,y)$ is $C^{\infty}$ in a possibly dependent on choice of $x$ neighborhood of $y_0$ and that differentiation of $\int_{\mathcal{X}}g(x,y)d\nu(x)$ can be always done within the integral? If it still does not work, the special case I am interested in is the exponential family case, namely $g(x,y) = \exp \{y^T h(x)\}$ for $h : \mathcal{X} \to \mathbb{R}^n$. The question asks basically if a function is integrable with respect to one element from the exponential family whose parameter is in the interior of the parameter space, is the function still integrable with respect to the parameters in a neighborhood of it.

By the given conditions, it suffices to show that, for any $y$ sufficiently close to $y_0$ :

\begin{equation*} \int_{\mathcal{X}} |f(x)||g(x,y)-g(x,y_0)|d\nu(x) < \infty \end{equation*}

It is, however, not clear how to proceed as the difference bounds are not uniform in $x$ and the space is infinite. Since the result is trivial for bounded $f$, I try truncating $f$, but it does not seem to be working.

Answer 1

Let $\mu$ be standard Gaussian measure on the real line and define $g(x,y)=e^{yx}y$ $$\int_{\mathbb{R}} g(x,y)\,\mu(dx)=y\int_\mathbb{R} e^{xy}e^{-x^2/2}\,dx<\infty$$ for all $y$. For $f(x)=e^{x^2/2}$, have that $$\int_{\mathbb{R}}|f(x)||g(x,0)|\mu(dx)<\infty$$ however, $|g(x,y)||f(x)|$ is not integrable (in $x$) for any other value of $y$.

---

Let's consider the case in which the OP interested, that is, suppose that $h$ is measurable with values in $\mathbb{R}^d$, $f$ is real measurable, $\mu$ is a $\sigma$-finite measure, and $$\int_{\mathcal{X}}e^{y\cdot h(x)}f(x)\mu(dx)<\infty$$ for some $y$. For simplicity, assume $f(x)\geq 0$. By Hölder's inequality the set $$E=\left\{y\in\mathbb{R}^d: \int_{\mathcal{X}}e^{y\cdot h(x)}f(x)\mu(dx)<\infty\right\}$$ is convex: Let $\nu(dx)=f(x)\,\mu(dx)$. If $y_1,y_2\in E$ and $0<t<1$, then \begin{align} \int_\mathcal{X} e^{(ty_1+(1-t)y_2)\cdot h(x)}\nu(dx)&=\int_\mathcal{X} e^{ty_1\cdot h(x)}e^{(1-t)y_2\cdot h(x)}\nu(dx)\leq\Big\|e^{ty_1\cdot h}\Big\|_{\tfrac1t}\Big\|e^{(1-t)y_2\cdot h}\Big\|_{\tfrac{1}{1-t}}\\ &=\Big(\int_{\mathcal{X}}e^{y_1\cdot h(x)}\,\nu(dx)\Big)^t\Big(\int_{\mathcal{X}}e^{y_2\cdot h(x)}\,\nu(dx)\Big)^{1-t}<\infty \end{align} If $E$ has no empty interior, then it can be shown that on $\operatorname{Int}(E)$, the map $\phi:y\mapsto \int_\mathcal{X} e^{y\cdot h(x)}\,\nu(dx)$ is analytic and that $$\partial_\alpha\phi(y)=\int_\mathcal{X}h^\alpha(x)e^{y\cdot h(x)}\,\nu(dx)$$ for any $\alpha\in\mathbb{Z}^d_+$. To see this, it is enough to consider the unidimensional case ($d=1$). Fix $y\in\operatorname{Int}(E)$ and choose $\delta>0$ so that $B(y;\delta)\subset \operatorname{Int}(E)$. For all $k$ with $|k|<\delta$ \begin{align} \frac{\phi(y+k)-\phi(y)}{k}=\int_{\mathcal{X}}\frac{e^{kh(x)}-1}{k}e^{yh(x)}\,\nu(dx) \end{align} The convexity of the exponential function implies that $$\Big|\frac{e^{kh(x)}-1}{k}\Big|\leq \frac{e^{|kh(x)|}-1}{|k|}\leq \frac{e^{\delta|h(x)|}-1}{\delta}\leq \frac{e^{\delta h(x|}+e^{-\delta h(x)}}{\delta} $$ The conclusion follows by dominated convergence.