Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.
(2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $\int_X f = \int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E=\{x : f(x) \neq g(x)\}$ in this case, then
$$\int_Xf = \int_{X \setminus E}f + \int_Ef = \int_{X \setminus E}f = \int_{X \setminus E}g = \int_Xg$$
(3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if
$$\sup_{\phi \leqslant f} \int_X \phi = \inf_{\psi \geqslant f}\int_X \psi,$$
where $\phi, \psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.
We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x \mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.
The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.
Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $\int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get
$$\int_X f = \lim_{n \to \infty} \int_X f_n$$
When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).
It depends on your definition of Lebesgue Integrable...
If $\int \delta d \lambda := \lim_{n \to \infty} \int \delta_n d \lambda,$ then, since the limit exists and is 1, $\delta$ is integrable.
However, usually the Lebesgue integral is defined as $$\int \delta d \lambda := \sup\left\{ \sum_{i=1}^n (b_i-a_i)\alpha_i \right\},$$ where the $\sup$ runs over all functions $\varphi(t) := \sum_{i=1}^n \alpha_i \cdot 1_{(a_i,b_i)}(t),$ that satisfy $0 \leq \varphi(t) \leq \delta(t).$ (Here $1_S(t)$ is the indicator function of the set $S$.) In this definition, $\int \delta d \lambda = 0$, so $\delta$ is integrable for a stupid reason: it's 0 almost everywhere.
PS: A generally difficult problem in analysis is to identify when $\lim \int f_n = \int \lim f_n.$ With your $\delta$ and $\delta_n$ functions, you have an example of when the two sides are not equal.
Best Answer
The sequence
$$f_n(x) = \sum_{k=1}^n \sin kx \, \mathbf{1}_{\left(2^{-k},2^{-k+1} \right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < \sin kx < 1$ for $x \in \left(2^{-k},2^{-k+1} \right]$ we have $|f_n(x)| \leqslant 1$ for all $x \in [0,1]$. By the LDCT, $f$ is integrable and
$$\int_{[0,1]} f \, d\mu= \lim_{n\to \infty}\int_{[0,1]} f_n \, d\mu = \lim_{n\to \infty}\sum_{k=1}^n \int_{(2^{-k},2^{-k+1}]} \sin kx \, d\mu \\ = \sum_{k=1}^\infty \int_{(2^{-k},2^{-k+1}]} \sin kx \, d\mu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$\int_{[0,1]} f \, d\mu = \sum_{k=1}^\infty \int_{2^{-k}}^{2^{-k+1}} \sin kx \, dx \\ = \sum_{k=1}^\infty \frac{\cos \frac{k}{2^{k}} - \cos \frac{k}{2^{k-1}}}{k} \\ \underbrace{\approx 0.605359}_{\text{WolframAlpha}}$$