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Let $\mu$ be standard Gaussian measure on the real line and define $g(x,y)=e^{yx}y$ $$\int_{\mathbb{R}} g(x,y)\,\mu(dx)=\frac{1}{\sqrt{2\pi}}y\int_\mathbb{R} e^{xy}e^{-x^2/2}\,dx<\infty$$ 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 $y$. This shows that it is possible to have a smooth function $g(x,y)$ and a $\sigma$-finite measure $\mu$ such that $\int_{\mathcal{R}} g(x,y)\,\mu(dx)$ is defined for all $y\in\mathbb{R}$, and a function $f(x)$ such that $\int_{\mathbb{R}}|g(x,y) f(x)|\,\mu(dx)<\infty$ only at a point $y=y_0$.

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.

The following examples show that that it is possible to have a smooth function $g(x,y)$ and a $\sigma$-finite measure $\mu$ such that $\int_{\mathcal{R}} g(x,y)\,\mu(dx)$ is defined for all $y\in\mathbb{R}$, and a function $f(x)$ such that $\int_{\mathbb{R}}|g(x,y) f(x)|\,\mu(dx)<\infty$ only at a point $y=y_0$.

1. Let $\mu$ be standard Gaussian measure on the real line and define $g(x,y)=e^{yx}y$ $$\int_{\mathbb{R}} g(x,y)\,\mu(dx)=\frac{1}{\sqrt{2\pi}}y\int_\mathbb{R} e^{xy}e^{-x^2/2}\,dx<\infty$$ 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 $y$.

2. Let $\mu=\big(\frac{1}{\sqrt{x}}\mathbb{1}_{\{|x|\leq1\}} + \frac{e^{-2|x|}}{x^2}\mathbb{1}_{\{|x|>1\}}\Big)\,dx$. Define $g(x,y)=e^{xy}$ and $f(x)=1_{\{|x|\leq1\}}+e^{2|x|}\mathbb{1}_{\{|x|>1\}}$. Observe that $$\int_\mathbb{R}|g(x,y)|\,\mu(dx)=\int_{|x|\leq1}e^{xy}\frac{dx}{\sqrt{x}} +\int_{|x|>1}e^{xy -2|x|} \frac{dx}{x^2} $$ is well defined for $|y|<2$; however, $$ \int_\mathbb{R}|g(x,y) f(x)|\,\mu(dx)=\int_{|x|\leq1}e^{xy}\frac{dx}{\sqrt{x}} +\int_{|x|>1}e^{xy} \frac{dx}{x^2} $$ is defined iff $y=0$.

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.

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.

Let $\mu$ be standard Gaussian measure on the real line and define $g(x,y)=e^{yx}y$ $$\int_{\mathbb{R}} g(x,y)\,\mu(dx)=\frac{1}{\sqrt{2\pi}}y\int_\mathbb{R} e^{xy}e^{-x^2/2}\,dx<\infty$$ 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 $y$. This shows that it is possible to have a smooth function $g(x,y)$ and a $\sigma$-finite measure $\mu$ such that $\int_{\mathcal{R}} g(x,y)\,\mu(dx)$ is defined for all $y\in\mathbb{R}$, and a function $f(x)$ such that $\int_{\mathbb{R}}|g(x,y) f(x)|\,\mu(dx)<\infty$ only at a point $y=y_0$.

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.

I just consider the case you are interested in, that is $$\int_{\mathcal{X}}e^{y\cdot h(x)}f(x)\mu(dx)<\infty$$ 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.

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.