# If distribution's mean is increasing, Is positive increasing function's expectation increasing?

Let me assume that the variable $$x\in [a,b]$$ and $$y\in[a,b]$$ where $$a,b (a are real numbers. I have positive increasing function $$g:[a,b] \rightarrow [c,d]$$, where $$c,d(c are real numbers. $$g(x') \ge g(x) \text{ and } g(x)\ge 0$$ for all $$x'\ge x$$.

And I have a pdf $$f:[a,b]\times [a,b] \rightarrow [0,1]$$. More specifically, $$f(x,y) = \frac{\frac{1}{\sqrt{2\pi \sigma^2}}exp(-\frac{(x-y)^2}{2\sigma^2})}{\int_{a}^{b}\frac{1}{\sqrt{2\pi \sigma^2}}exp(-\frac{(x-y)^2}{2\sigma^2})dx},$$ which means the boosted $$y$$-mean Gaussian distribution for $$x$$ on the interval $$[a,b]$$.

I want to prove, for fixed $$x$$, $$\int_{a}^{b}g(x)f(x,y)dx$$ is increasing for $$y$$ or not. I think it should be increasing, but don't know how to prove this. Actually, I want to find a pdf which makes the last integral form increase by using Normal distribution.