The problem
Let $X$ be uniformly distributed on interval $[500,600]$ and $Y$ be uniformly distributed on interval $[400,800]$. We also define $$Z:=\min\{X,Y\}$$ i want to find $E(Z|Y)$ and with using it, I want to count $E(Z)$.
My work so far
Let's start with $E(\min\{X,Y\}|Y)$.
For $Y \in [400,500]$ we have $\min\{X,Y\}=Y \Rightarrow E(Z|Y)=E(Y|Y)=Y$
For $Y \in [600,800]$ we have $\min\{X,Y\}=X \Rightarrow E(Z|Y)=E(X|Y)=E(X)=550$.
For $Y \in [500,600]$ we can't tell nothing about our minimum, so I stuck in that point.
$$E(Z)=E(Z|Y \in [400,500]) \cdot P(Y \in [400,500])+E(Z|Y \in [500,600]) \cdot P(Y \in [500,600])+E(Z|Y \in [600,800]) \cdot P(Y \in [600,800])$$.
The Question
Can you please with counting $E(Z|Y)$ when $Y$ is on interval $[500,600]$ ?.
Also, I have a question if my pattern which is counting $E(Z)$ is correct.
You can directly compute the expectation of $Z$ as follows: \begin{align*} \mathbb E[\min(X,Y)]&=\int_{400}^{800}\int_{500}^{600}\min(x,y)\dfrac{\mathrm{d} x}{100}\dfrac{\mathrm{d} y}{400}\\ &=\dfrac{1}{40000}\left(\int_{400}^{500}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y + \int_{500}^{600}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y + \int_{600}^{800}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y\right) \\ &=\dfrac{1}{40000}\left(\int_{400}^{500} 100y\ \mathrm{d}y +\int_{500}^{600}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y+ \int_{600}^{800} \dfrac{600^2-500^2}{2}\ \mathrm{d}y\right) \\ &=\dfrac{1}{40000}\left(\dfrac{1}{2}\left(100(500^2-400^2)+200(600^2-500^2)\right)+\int_{500}^{600}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y\right) \\ &=387.5+\dfrac{1}{40000}\left(\int_{500}^{600}\int_{500}^{600}\min(x,y) dxdy\right)\\ &=387.5+\dfrac{1}{40000}\left(\int_{500}^{600}\left(\int_{500}^{y}\min(x,y) \ \mathrm{d}x + \int_y^{600} \min(x,y)\ \mathrm{d}x\right)\ \mathrm{d}y\right)\\ &=387.5+\dfrac{1}{40000}\left(\int_{500}^{600}\left(\int_{500}^{y}x \ \mathrm{d}x + \int_y^{600} y\ \mathrm{d}x\right)\ \mathrm{d}y\right)\\ &=387.5+\dfrac{1}{40000}\left(\int_{500}^{600}\left(y^2/2-125000 + (600-y) y\right)\ \mathrm{d}y\right)\\ &=387.5+\dfrac{1}{40000}\left(16000000/3\right)\\ &=387.5+400/3\\ &=\dfrac{3125}{6}\approx 520.83. \end{align*}
As far as $\mathbb E[\min(X,Y)\ | Y]$ is concerned, you can obtained it as the inner integral: \begin{align*} \mathbb E[\min(X,Y)\ | Y] &=\int_{500}^{600}\min(x,y)\dfrac{\mathrm{d} x}{100}\\ &=\dfrac{1}{100}\left(100y \ 1_{y \leq 500} + \int_{500}^{600} \min(x,y) \ 1_{500\leq y\leq 600} \ \mathrm{d}y+55000 \ 1_{y\geq 600} \right)\\ &=y \ 1_{y \leq 500} + 550 \ 1_{y\geq 600} + \dfrac{1}{100}\left(\int_{500}^{y}\min(x,y) \ \mathrm{d}x + \int_y^{600} \min(x,y)\ \mathrm{d}x\right)\\ &=y \ 1_{y \leq 500} + 550 \ 1_{y\geq 600} + \dfrac{1}{100}\left(y^2/2-125000 + (600-y) y\right)\\ &=y \ 1_{y \leq 500} + 550 \ 1_{y\geq 600} + \dfrac{1}{100}\left(-y^2/2-125000 + 600 y\right). \end{align*} Integrating this quantity with respect to $y$ gives the previous result.