$\sigma$-algebra generated by random variable : Show that if $\sigma(X)=\sigma(Y)$ then $\sigma(X+Y)\subseteq \sigma(X)$ Let $(\Omega,\mathcal{F},P)$ be a probability space and $X$ be a random variable. The $\sigma$-algebra generated by $X$ is defined as
$$\sigma(X):=\{X^{-1}(B)\; | \; B\in B_{\mathbb{R}}\}$$
where $B_{\mathbb{R}}$ is the Borel $\sigma$-algebra of subsets of $\mathbb{R}$. The problem statement is as follows :
Let $(\Omega,\mathcal{F},P)$ be a probability space and $X,Y$ are random variables such that $\sigma(X)=\sigma(Y)$. Show that $\sigma(X+Y)\subseteq \sigma(X)$.
I can show that $X+Y$ is a random variable. So, how do I go about proving this?

By definition, $\sigma(Z)$ denotes the smallest $\sigma$-algebra $\Sigma$ on $\Omega$ such that $Z: (\Omega, \Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable.
This means in particular that $X:(\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable. Since $\sigma(Y) = \sigma(X)$, we also have that
$$Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable. Consequently, the sum
$$X+Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable as a sum of two measurable random variables. Since $\sigma(X+Y)$ is the smallest $\sigma$-algebra $\Sigma$ such that
$$X+Y: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable, this shows $\sigma(X+Y) \subseteq \sigma(X)$.
