For $x\in \mathbb{R}$, let $u(x,t) = (\Gamma_{D} (\cdot , t ) \ast g)(x)$ with $g(x) \ge 0,$ and $\int_{-\infty}^{\infty}g(x) dx < \infty$. Prove that for all $t>0$, $$ \int_{-\infty}^{\infty} u(x,t) dx = \int_{-\infty}^{\infty} g(x) dx,$$ the meaning of this statement is that the total heat is conserved. Using the formula for $u(x,t)$, show by direct integration that for all $t > 0$, $$ \int_{-\infty}^{\infty} u(x,t) \ dx = \int_{-\infty}^{\infty} g(x) \ dx.$$ Hint: Because $g(x) \ge 0$, and $\Gamma_{D} (x,t) \ge 0$ , you may always exchange the order of iterated integrals
Attempt at solution
Since by proprety $$ \int_{-\infty}^{\infty}\Gamma(x,t) \ dx =1,$$ then it follows by change of variable that $$ \int_{-\infty}^{\infty}\Gamma(x-y,t) \ dx =1$$ as well. So then, \begin{align} \int_{-\infty}^{\infty} u(x,t) \ dx &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Gamma(x-y,t)g(y) \ dy \ dx \\ &= \int_{-\infty}^{\infty} g(y) \int_{-\infty}^{\infty} \Gamma(x-y,t) \ dx \ dy \\ &= \int_{-\infty}^{\infty} g(y) \ dy \end{align} Since $y$ is a dummy variable this is equivalent to $$ \int_{-\infty}^{\infty} u(x,t) \ dx = \int_{-\infty}^{\infty} g(x) \ dx.$$
I'm not sure if my proof holds, particularly I'm not sure if what I just provided falls into the category of "direct integration".