Prove that $\sigma_{\text {ess}} (A)$ is a closed subset of $\mathbb R.$ Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal H.$ Let $E_A$ be the unique spectral measure associated to $A$ obtained from spectral theory for self-adjoint operators defined on the Borel-$\sigma$-algebra of subsets of $\left [-\|A\|, \|A\| \right ]$ i.e.
$$A = \int_{\left [-\|A\|, \|A\| \right ]} t\ dE_A(t).$$
Let $\sigma (A)$ denote the spectrum of $A.$ Then what I know is that for any self-adjoint operator $A$ on a Hilbert space $\mathcal H$ we have $\sigma (A) = \text {supp} (E_A),$ where $\text {supp} (E_A)$ denotes the support of $E_A.$ This shows that if $\lambda \in \sigma (A)$ then $E_A (\lambda - \varepsilon, \lambda + \varepsilon) \neq 0,$ for every $\varepsilon \gt 0.$ This leads us to the following subdivision of the spectrum $\sigma (A)$ of $A.$
An element $\lambda \in \sigma (A)$ is said to be an essential spectrum of $A$ if the range of the projection $E_A (\lambda - \varepsilon, \lambda + \varepsilon)$ is infinite dimensional for every $\varepsilon \gt 0.$ Otherwise we say that $\lambda$ is a discrete spectrum of $A.$ The collection of all essential spectrum of $A$ is denoted by $\sigma_{\text {ess}} (A)$ and the collection of all discrete spectrum of $A$ is denoted by $\sigma_{\text {disc}} (A).$
Now two results have been left as (easy) exercises which are the following $:$

$(1)$ $\sigma_{\text {ess}} (A)$ is a closed subset of $\mathbb R$ for any self-adjoint operator $A$ on a Hilbert space $\mathcal H.$
$(2)$ If $\lambda \in \sigma_p(A)$ has infinite multiplicity then $\lambda \in \sigma_{\text {ess}} (A),$ where $\sigma_p (A)$ denotes the point spectrum (or the collection of eigenvalues) of $A.$

But I find it difficult to prove the first one. I have tried by taking a sequence $\{\lambda_n\}_{n \geq 1}$ in $\sigma_{\text {ess}} (A)$ converging to $\lambda.$ Then the range of the projection $E_A (\lambda_n - \varepsilon, \lambda_n + \varepsilon)$ is infinite dimensional for every $\varepsilon \gt 0$ and for all $n \geq 1.$ But how does it guarantee that the range of the projection $E_A (\lambda - \varepsilon, \lambda + \varepsilon)$ is also infinite dimensional for all $\varepsilon \gt 0\ $? I have asked about it to our instructor. He told me that it is an one line argument. But I don't know why can't I able to see the proof. Also I don't have any idea about the second one. May be I am so stupid. Would anybody give me some suggestion here? I am totally confused at thus stage about how to proceed further.
Any help regarding this will be warmly appreciated. Thanks!
EDIT $:$ Finally I am able to prove the first one. Let us take $\varepsilon \gt 0$ arbitrarily. Let $\{\lambda_n\}_{n \geq 1}$ be a sequence in $\sigma_{\text {ess}} (A)$ converging to $\lambda \in \mathbb R.$ So there exists $N \geq 1$ such that $\lambda_n \in (\lambda - \varepsilon, \lambda + \varepsilon),$ for all $n \geq N.$ In particular $\lambda_N \in (\lambda - \varepsilon, \lambda + \varepsilon).$ Choose $\delta \gt 0$ small enough so that $(\lambda_N - \delta, \lambda_N + \delta) \subseteq (\lambda - \varepsilon, \lambda + \varepsilon).$ This implies that $E_A ((\lambda_N - \delta, \lambda_N + \delta)) \leq E_A ((\lambda - \varepsilon, \lambda + \varepsilon)).$ But this in turn implies that $$\text {Range} \left (E_A ((\lambda_N - \delta, \lambda_N + \delta)) \right ) \subseteq \text {Range} \left ( E_A ((\lambda - \varepsilon, \lambda + \varepsilon)) \right ).$$ Now since $\lambda_N \in \sigma_{\text {ess}} (A)$ it follows that $\text {Range} \left (E_A ((\lambda_N - \delta, \lambda_N + \delta)) \right )$ is infinite dimensional and hence we have $\text {Range} \left (E_A ((\lambda - \varepsilon, \lambda + \varepsilon)) \right )$ is infinite dimensional. This completes the proof.
Now how do I prove the second one? Do anybody give any idea about it? Thanks!
 A: Q2. $\lambda \in \sigma(A)$ is an eigenvalue if and only if $E_A(\{\lambda\})\neq 0$. In this situation, $E_A(\{\lambda\})$ is the projection onto the eigenspace with respect to $A$ at the point $\lambda$. If the eigenvalue $\lambda$ has infinite multiplicty, then $E_A((\lambda - \varepsilon, \lambda + \varepsilon))$ must have infinite rank for every $\varepsilon>0$.
A: Your comments above are how I would proceed, but I don't follow how you arrived at the important inclusion for this problem: $\ker(A-\lambda)\subseteq \text{Ran}E_{A}(\lambda)$. For example, how does $v$ not in the range imply $E_{A}(\lambda)v=0$? It may have a component in the range and one in the orthogonal complement. Then I also don't follow how you conclude that the measure should concentrate around $\lambda$. The other inclusion looks good, however.
Here's how I would prove the inclusion $\ker(A-\lambda)\subseteq \text{Ran}E_{A}(\lambda)$. For $v\in \ker(A-\lambda)$, then $d\mu_v=\| v\|^2\delta_\lambda$. Thus,
$$
0=\langle v,E_{A}(\mathbb{R}\setminus\{ \lambda\})v\rangle
$$
and by orthogonality of the projection, $v\in \ker(E_{A}(\mathbb{R}\setminus \{\lambda\})=\text{Ran}(I-E_{A}(\mathbb{R}\setminus \{\lambda\}))=\text{Ran}(E_A(\lambda))$.
So in particular, if $\dim(\ker(A-\lambda))=+\infty$, then $\dim(\text{Ran}(A-\lambda))=+\infty$ and $\lambda\in \sigma_{ess}(A)$.
edit: for future readers, OP had attempted a proof that $\ker(A-\lambda)=E_{A}(\lambda)$ in the comments. While this is true, the proof of one of the inclusions was incorrect (and is the important one for solving this problem). I have provided a proof of this above.
