Why $\mathbb P\{X+Y\leq t\}=\mathbb P\{X\leq t-y,Y\leq t\}$? Let $X,Y$ two continuous r.v.
Why $$\mathbb P\{X+Y\leq t\}=\mathbb P\{X\leq t-y,Y\leq y\}\ \ ?$$
To me, it would be $\mathbb P\{X\leq t-y,Y=y\}$. 
 A: The probability in question is a region in the plane such that $y\leq t-x$. Your interpreted region lies entirely on the line $Y=y$. This probability will be zero much of the time! (for example take $X$, $Y$ independent.)
A: What you've written down isn't correct. For example, if you let $X$ and $Y$ be trivial random variables which always take on the value 1, then for $t = 2$, the probability on the RHS is 1. Choosing $y = 2$, the probability on the RHS is 0.
Perhaps what you mean to write is
\begin{align*}
P(X + Y \leq t) = \int_{-\infty}^t f_{X+Y}(z) dz
\end{align*}
where $f_{X+Y}(z)$ is the density of the continuous random variable $X + Y$. This is given by
\begin{align*}
f_{X+Y}(z) = \int_{\mathbb{R}} f_X(w) f_Y(z - w) dw
\end{align*}
So the expression becomes
\begin{align*}
P(X + Y \leq t) = \int_{-\infty}^t \int_{\mathbb{R}} f_X(w) f_Y(z - w) dw dz
\end{align*}
Since the integrand is nonnegative, we can switch the order of integration to get the above is equal to
\begin{align*}
&= \int_{\mathbb{R}} f_X(w) \int_{-\infty}^t f_Y(z-w) dz dw 
\end{align*}
Substituting $u = z-w$, we have
\begin{align*}
&= \int_{\mathbb{R}} f_X(w) \int_{-\infty}^{t-w} f_Y(u) du dw \\
&= \int_{\mathbb{R}} f_X(w) P(Y \leq t - w) dw \\
\end{align*}
In conclusion, your intuition is almost correct, except that you would then need to integrate over all $y$, and replace the probability by a probability density function.
