Proving the identity $P( X + Y = a)= \int_{-\infty}^{\infty} P( X + y = a)f_Y(y) \, \text{d}y $ Suppose $\lambda_1, \lambda_2, a \in \mathbb{R}$ and $X,Y$ are random variables. If it is needed, I can assume that $X$ and $Y$ are independent. I want to show, that the identity
$$\mathbb{P}(\lambda_1 X + \lambda_2 Y = a)= \int_{-\infty}^{\infty} \mathbb{P}(\lambda_1 X + \lambda_2 y = a)f_Y(y) \, \text{d}y $$
does hold, where $f_Y$ denotes the density function of $Y$.
The idea behind this identity is that it is very useful, when the density $f_Y$ is given and we know something about the probability $\mathbb{P}(\lambda_1 X + \lambda_2 y = a)$. 
Now I want to prove this statement. I tried to use the substitution formula
$$\mathbb{E}[g(X)]= \int_{\Omega}g(X(\omega)) \, \text{d} \mathbb{P}(\omega) = \int_{\mathbb{R}} g(x) \, \text{d}P_X(x) = \int_{-\infty}^{\infty} g(x) f_X(x) \, \text{d}x, $$
where $g: \mathbb{R} \rightarrow \mathbb{R}$ is measurable and calculated the following:
$$\mathbb{P}(\lambda_1 X + \lambda_2 Y = a)= \mathbb{E}[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}]= \int_{\Omega}\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}(\omega) \, \text{d} \mathbb{P}(\omega)= \int_{-\infty}^{\infty} ..?.. \, \text{d} P_{(X,Y)}(x,y),$$
but I do not know how to continue at this point.
 A: Use the Law of Iterated Expectation.
$$\begin{align}
\mathbb{P}(\lambda_1 X + \lambda_2 Y = a) & = \mathbb{E}[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}]
\\[2ex] & = \Bbb E[\Bbb E[\mathbf 1_{\lambda_1 X + \lambda_2 Y = a}\mid Y]]
\\[2ex] & = \int_{\Bbb R}\Bbb E[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}\mid Y](y) \, \text{d} \mathbb{P}_Y(y)
\\[2ex] & = \int_{-\infty}^{\infty} \Bbb P(\lambda_1X+\lambda_2 y=a\mid Y=y)\, f_Y(y) \, \mathrm{d} y
\end{align}$$
Note: only if X and Y are independent does this equal $\int_{-\infty}^{\infty} \Bbb P(\lambda_1X+\lambda_2 y=a) \,f_Y(y) \mathrm{d} y$
A: Since the original poster wasn't satified with a step in Grahm's answer I will add the following clarification of that step.

How do we motivate that $\int P[\lambda_1X+ \lambda_2Y=a|Y=y]P_Y(dy) $ equals $\int P[\lambda_1X+\lambda_2y=a]P_Y(dy)$?

This follows from the following result which may be found on p.263 in Shiryaev's Probability 1 (2016)

If $X,Y $ are independent and $\phi $ is a $\mathcal B(R^2) $
  measurable function such that $E[|\phi(X,Y) |]<\infty $ then  $$\int E[\phi(X,Y)|Y=y]P_Y(dy)=\int E[\phi(X,y)]P_Y(dy)$$

We simply let $\phi(x,y)=1_{\{x+y=a\} }(x,y)$. Hope this helps!
