Conditional distribution function of one random variable given the sum of two I am trying to solve the following exercise in Probability Theory by A. Klenke (3rd version).
Let X and Y be independent exponential random variables for some $\theta>0$. Compute $P[X \leq x | X+Y]$ for $x\geq0$.
My solution is based only on the definition of conditional expectation and in particular on this property: if $\mathbb{E}[\mathbb{1}_A X]=\mathbb{E}[\mathbb{1}_A\mathbb{E}[X|\mathcal{F}]]$ for every $A \in \mathcal{F}$ then $\mathbb{E}[X|\mathcal{F}]$ is called a conditional expectation, where $X\in\mathcal{L}^1(\Omega, \mathcal{A},\mathbb{P})$ and $\mathcal{F}\subset \mathcal{A}$ are two $\sigma$-algebras.
Thus, for every $A\in \sigma(X+Y)$:
$\int_A \mathbb{1}_{X(\omega)\in[0,x]} d\mathbb{P}=\int_A\mathbb{1}_{X(\omega)\in[0,x]} d(\mathbb{P}\circ(X \times (X+Y))^{-1})=\int_A\int_0^t\mathbb{1}_{t-y\in[0,x]} \theta e^{-\theta(t-y)}\theta e^{-\theta y}dydt=\int_A \int_{t-x}^{t} \theta^2e^{-\theta t}dydt=\int_A \frac{x}{t} t\theta^2e^{-\theta t}dt = \int_A \frac{x}{T} d\mathbb{P}$.
So I conclude: $P[X \leq x | X+Y] = \frac{x}{X+Y}$.
In the second equality I obtained the density of $(X,T)$, where $T=X+Y$, in this way: $f_{X,T}(x,t)=f_{X,Y}(t-y,y)=f_X(t-y)f_Y(y)$ by the independence property.
Is this correct?
Edit
Taking in the comments made by @D Ford, if I define $T=X+Y$, then this is the correct chain of equalities:
$\int_A \mathbb{1}_{X\in[0,x]}(\omega) d\mathbb{P}= \\
\int_{T(A)}\mathbb{1}_{[0,x]}(X) d(\mathbb{P}\circ(X \times (X+Y))^{-1})=\\
\int_{T(A)}\int_0^t\mathbb{1}_{t-y\in[0,x]} \theta e^{-\theta(t-y)}\theta e^{-\theta y}dydt=\\
\int_{T(A)} \int_{t-x}^{t} \theta^2e^{-\theta t}dydt=\\
\int_{T(A)} \frac{x}{t} t\theta^2e^{-\theta t}dt = \\
\int_A \frac{x}{T} d\mathbb{P}$.
 A: 
In the second equality I obtained the density of (X,T), where T=X+Y, in this way: $f_{X,T}(x,t)=f_{X,Y}(t−y,y)=f_X(t−y)f_Y(y)$ by the independence property.
Is this correct?

No. You have the right idea, but you begin with a function of $x$ and $t$, so should not end with a function of $t$ and $y$.
Rather:
$$\begin{align}f_{X,T}(x,t)&=f_{X,Y}(x,t-x)\\&=f_X(x)\cdot f_Y(t-x)\\&= \theta^2\mathrm e^{-\theta x}\mathrm e^{-\theta (t-x)}\mathbf 1_{0\leq x}\mathbf 1_{0\leq t-x}\\&=\theta^2\mathrm e^{-\theta t}\,\mathbf 1_{0\leq x\leq t}\end{align}$$

And similarly we might obtain the same result:.
$$\begin{align}\mathsf P(X\leq x\mid X+Y=t) &=\dfrac{\int_0^x f_{X,Y}(s,t-s)\,\mathrm d s}{\int_0^t f_{X,Y}(s,t-s)\,\mathrm d s}\mathbf 1_{0\leq x\lt t}+\mathbf 1_{t\leq x}\\[2ex]&=\dfrac{\theta^2\mathrm e^{-\theta t}\int_0^x \mathrm ds }{\theta^2\mathrm e^{-\theta t}\int_0^t \mathrm ds}\,\mathbf 1_{0\leq x\lt t}+\mathbf 1_{t\leq x}\\[2ex]&=\dfrac{x}{t}\,\mathbf 1_{0\leq x<t}+\mathbf 1_{t\leq x}\end{align}$$

So I conclude: $P[X≤x\mid X+Y]=x/(X+Y)$.

A: Here's another solution to this problem. If we know $g : \mathbb R^2 \to \mathbb R$ is continuous (as we suspect the joint density of $X$ and $X+Y$ to be), then we can use the fundamental theorem of calculus:
$$
g(x,y) = \frac{\partial^2}{\partial x \partial y} \int_0^x \int_0^y g(s,t) \, ds \, dt. 
$$
In this case, to find the joint density of $X$ and $X+Y$, we first observe:
$$
\mathbb 1_{\{X \leq x\} \cap \{X+Y \leq z\}} = \mathbb 1_{\{X \leq x \} \cap \{Y \leq z - X\}} = \mathbb 1_{A(x,z)}(X,Y),
$$
where $A(x,z) = \{(s,t) \in \mathbb R^2 : 0 \leq s \leq x, 0 \leq z \leq y-s\}$.
So we compute:
\begin{align*}
\mathbb P \left[ \{X \leq x\} \cap \{X+Y \leq z\}\right] &= \int \mathbb 1_{A(x,z)}(X,Y) \, d\mathbb P \\
&= \int_{A(x,z)} d\left(\mathbb P \circ(X \times Y)^{-1}\right) \\
&= \int_0^x \int_0^{z-s} \theta^2 e^{-\theta(s+t)} \, dt \, ds \\
&= 1 - e^{-\theta x} - \theta x e^{-\theta z}.
\end{align*}
Differentiating this with respect to $x$ and $z$, and noting $X, Y \geq 0$ and $\mathbb P[\{X > z\} \cap \{X+Y \leq z\}] = 0$, we find that the joint density $f$ of $X$ and $X+Y$ is
$$
f(x,z) = \theta^2 e^{-\theta z} \mathbb 1_{[x,\infty)}(z) \mathbb 1_{[0,\infty)}(x).
$$
This joint density, along with part (i) of this exercise, can be used to compute both $\mathbf E[X|X+Y]$ and $\mathbf P[X \leq x | X+Y]$.
A: There are a couple issues in your new chain of equalities that should be addressed, and I felt it was probably better to articulate them as an answer rather than post a comment.

[T]his is the correct chain of equalities:
$\int_A \mathbb{1}_{X\in[0,x]}(\omega) d\mathbb{P}= \\
\int_{T(A)}\mathbb{1}_{[0,x]}(X) d(\mathbb{P}\circ(X \times (X+Y))^{-1})=\\
\int_{T(A)}\int_0^t\mathbb{1}_{t-y\in[0,x]} \theta e^{-\theta(t-y)}\theta e^{-\theta y}dydt=\\
\int_{T(A)} \int_{t-x}^{t} \theta^2e^{-\theta t}dydt=\\
\int_{T(A)} \frac{x}{t} t\theta^2e^{-\theta t}dt = \\
\int_A \frac{x}{T} d\mathbb{P}$.

I agree with the second, third, and fourth. The fifth is correct, but needs some justification: why do you know the density of $T = X+Y$ is $t\theta^2 e^{-\theta t}$? That’s essentially what the final equality is saying. The second works after making the change of variables $X = t-y$ and using what we know about the joint density of $(X, X+Y)$.
The first equality, however, still needs some work.
The integral $\int_E f \, d\mu$ has three objects: a measure space $E$, a measure $\mu$ on $E$, and a (real or complex)-valued function $f$ defined on $E$. Your first integral is $\int_A \mathbb 1_{X \in [0,x]}(\omega) d\mathbb P$; here, $E = A$, $\mu=\mathbb P$ (a measure defined on $A$), and $f = \mathbb 1_{X \in [0,x]}$ (a function defined on $A$). So this works. (Although usually if it can be avoided we don’t include the independent variable in the integrand, $\omega$ in your case.)
Now consider your second integral: $\int_{T(A)} \mathbb 1_{[0,x]}(X) d(\mathbb P(X \times (X+Y))^{-1})$. Here, the space is $E = T(A)$ (a subset of $\mathbb R$), the function is $f = \mathbb 1_{[0,x]}(X) = \mathbb 1_{[0,x]} \circ X$ (a composition of functions whose domain is $\Omega$, not $\mathbb R$), and the measure is $\mu = P \circ (X \times (X+Y))^{-1}$. This measure is defined on $\mathbb R^2$, because the codomain of the function $X \times (X+Y)$ is $\mathbb R^2$, so if $B$ is a subset, then $\mathbb P \circ (X \times (X+Y))^{-1}(B) = \mathbb P((X \times (X+Y))^{-1}(B))$; but this expression only makes sense if $\mathbb B \subset \mathbb R^2$. So the measure isn’t defined on $T(A)$ (a subset of $\mathbb R$, not $\mathbb R^2$).
Now, the integral transformation law tells us that if $E$ and $F$ are measurable spaces, $\mu$ a measure on $E$, $T : E\to F$ a measurable map, and $f : F \to \mathbb R$ a real-valued function, then $$\int_E f \circ T d\mu = \int_F f d(\mu \circ T^{-1}).$$ This makes sense: $f \circ T$ is defined on $E$, so the first integral works, and $\mu \circ T^{-1}$ is a measure on $F$, so the second integral works. For example, if $Z$ is a random variable and $B \subset \mathbb R$ is measurable, then:
$$
\mathbb P[Z \in B] = \int \mathbb 1_{Z \in B} d\mathbb P = \int \mathbb 1_B(Z) d\mathbb P = \int \mathbb 1_B d(\mathbb P \circ Z^{-1}).
$$
Applying this to the first of your edited equalities, you start with $\int_A \mathbb 1_{X \in [0,x]} d\mathbb P$. If you want to integrate over $T(A)$, you would need to express the function $\mathbb 1_{X \in[0,x]}$ as a composition $f \circ T$ in the first integral to do that, for some function $f : \mathbb R \to \mathbb R$, but it’s not clear to me how you would do that. Instead, you want to integrate over $(X \times T)(A) = X(A) \times T(A)$, which it looks like you ultimately end up doing anyways. Then, we could say:
$$
\mathbb 1_{X \in [0,x]} = \mathbb 1_{[0,x] \times [0,\infty)}(X, X+Y)
$$
Or, better yet, if $A \in \sigma(X+Y)$, then $A = (X+Y)^{-1}(B)$ for some $B \in \mathbb R$, so:
$$
\mathbb 1_A \mathbb 1_{X \in [0,x]} = \mathbb 1_{[0,x] \times B}(X, X+Y)
$$
Implicitly, this last thing is what you end up doing anyways. But again, you want to justify your use of those densities.
