$f = 0$ almost everywhere implies Lebesgue integrability and $\int f = 0$ 
Let $f \in \mathcal{B}[a,b]$ so that $f = 0$ a.e. in $[a,b]$. Show that $f \in \mathcal{L}[a,b]$ and $\int_a^b f = 0$.

Proof. Since $f$ is bounded, then there is some $L \in \mathbb{R}$ such that $$\sup_{x\in[a,b]} |f(x)| < L.$$ Furthermore, since $f = 0$ almost everywhere then $m(\{ f \neq 0 \}) = 0$. Consider the upper and lower sums over a measurable partition $P = \{E_{j}\}_{j=1}^{n}$, i.e.
\begin{align*}
    U[f,P] = \sum_{j=1}^{n}M_j\cdot m(E_{j}) \\
    L[f,P] = \sum_{j=1}^{n}m_j\cdot m(E_{j})
\end{align*}
Now, consider only those $E_{k}$ where $f \neq 0$, call this $\widetilde{P} = \{E_{k}\}_{k=1}^{m}$, $m < n$. Note that $m(\widetilde{P}) = 0 \implies m(E_{k}) = 0$ for all $k=1,\ldots,m$, on account of the assumption. This then implies
$$ 0 = L[f,\widetilde{P}]  \leqslant U[f,\widetilde{P}] = 0 \implies L[f,\widetilde{P}] = U[f,\widetilde{P}] = 0$$
which implies that $f \in \mathcal{L}[a,b]$ and $\int_a^b f = 0$.
I'm looking for critiques of my proof. One thing I notice is that I haven't really used the boundedness.

If I interpreted the question correctly (see the comments), then I can only make the following remarks:
The Lebesgue measure is complete. This means that any subset of a null measurable set is also measurable and null.
From this, it should be straightforward to see that any $f : \mathbb R \to \mathbb R$ that is zero almost everywhere is measureable, whether $f$ is bounded or not. For instance, the preimage $f^{-1}(a,\infty)$ with $a > 0$ is a subset of the null set on which $f$ is zero, hence $f^{-1}(a,\infty)$ is itself measurable (and null). You can check the other preimages for yourself.
Secondly, if $f$ and $g$ are measurable functions with $f$ integrable and $f = g$ almost everywhere, then $g$ is also integrable and $\int g = \int f$. Essentially this is because any simple function that sits underneath $f$ also sits underneath $g$, once you "trim off" some null bits that do not affect the measurability or the measures of the domains on which the "blocks" are supported. The same is true with $f$ and $g$ reversed.
So it looks like the question is trivial. If anyone disagrees, please do leave a comment.
