No. In fact, every Lebesgue measurable function $f\colon I\to E$ is equal almost everywhere to a limit of simple Lebesgue measurable functions. As you hint at in the question, this is easy to show in the case where $E$ is separable. The general situation reduces to the separable case due to the following result. For a full proof, see Fremlin, *Measure Theory*, Volume 4 Part I, Lemma 451Q.

**Theorem 1:** Let $(X,\Sigma,\mu)$ be a finite compact measure space, $Y$ a metrizable space, and $f\colon X\to Y$ a measurable function. Then, there is a closed separable subspace $Y_0$ of $Y$ such that $f^{-1}(Y\setminus Y_0)$ is negligible.

That is, $f$ has essentially separable image. Restricting $f$ to the complement of a negligible set reduces the problem to the situation where the codomain is separable, in which case it is a limit of simple functions. Compactness of the space $(X,\Sigma,\mu)$ means that there is a family $\mathcal{K}\subseteq\Sigma$ such that any subset of $\mathcal{K}$ with the finite intersection property has nonempty intersection, and such that $\mu$ is inner-regular with respect to $\mathcal{K}$. That is, $\mu(E)=\sup\{\mu(K)\colon K\in\mathcal{K},K\subseteq E\}$ for every $E\in\Sigma$. In particular, the Lebesgue measure is compact by taking $\mathcal{K}$ to be the collection of compact sets under the standard topology.

The proof of Theorem 1 is rather tricky, involving what Fremlin describes as "non-trivial set theory". It rests on the following two results.

**Theorem 2:** Any metrizable space has a $\sigma$-disjoint base $\mathcal{U}$. That is, $\mathcal{U}$ is a base for the topology, and can be written as $\bigcup_{n=1}^\infty\mathcal{U}_n$ where each $\mathcal{U}_n$ is a disjoint collection of sets.

(Fremlin, *Measure Theory*, Volume 4 II, 4A2L (g-ii))

**Theorem 3:** Let $(X,\Sigma,\mu)$ be a finite compact measure space and $\{E_i\}_{i\in I}$ be a disjoint family of subsets of $X$ such that $\bigcup_{i\in J}E_i\in\Sigma$ for every $J\subseteq I$. Then, $\mu\left(\bigcup_{i\in I}E_i\right)=\sum_{i\in I}\mu(E_i)$.

(Fremlin, *Measure Theory*, Volume 4 I, 451P).

Theorem 3 is particularly remarkable, as it extends the countable additivity of the measure to arbitrarily large unions of sets.

Once these two results are known, the proof that $f$ has essentially separable image in Theorem 1 is straightforward. Let $\mathcal{U}=\bigcup_{n=1}^\infty\mathcal{U}_n$ be a $\sigma$-disjoint base for $Y$. Let $\mathcal{V}_n$ be the collection of $U\in\mathcal{U}_n$ such that $\mu(f^{-1}(U)) = 0$. By countable additivity, $\mathcal{U}_n\setminus\mathcal{V}_n$ is countable. Also, $\{f^{-1}(U)\colon U\in\mathcal{V}_n\}$ is a disjoint collection of negligible subsets of $X$ and, by measurability of $f$, any union of a subcollection of these is measurable. It follows from Theorem 3 that its union is negligible. That is, $f^{-1}\left(\bigcup\mathcal{V}_n\right)$ is negligible. Setting $Y_0=Y\setminus\bigcup_n\bigcup\mathcal{V}_n$ then, by countable additivity, $f^{-1}(Y\setminus Y_0)$ is negligible. Also, $\bigcup_n(\mathcal{U}_n\setminus\mathcal{V}_n)$ restricts to a countable base for the topology on $Y_0$, so it is separable (in fact, it is second-countable).

Finding a $\sigma$-disjoint base for the topology on $Y$ is easy enough. Following Fremlin, you can do this by well-ordering $Y$ and letting $(q_n,q^\prime_n)$ be a sequence running through the pairs $(q,q^\prime)$ of rationals with $0 < q < q^\prime$. Letting $\mathcal{U}_n$ be the collection of sets of the form
$$
G_{ny}=\left\{x\in Y\colon d(x,y) < q_n, \inf_{z < y}\,d(x,z) > q_n^\prime\right\}
$$
(over $y\in Y$) gives a $\sigma$-disjoint base.

The really involved part of the proof is in establishing Theorem 3. I suggest you look in Fremlin for the details, but the idea is as follows. By countable additivity, only countably many $E_i$ can have positive measure so, removing these, we can suppose that every $E_i$ is negligible. Also, restricting $X$ to the union of the $E_i$ if necessary, we can suppose that $X=\bigcup_iE_i$. Then define the function $f\colon X\to I$ by $f(x)=i$ for $x\in E_i$. Using the power set $\mathcal{P}I$ for the sigma-algebra on $I$, $f$ will be measurable. Then let $\nu=\mu\circ f^{-1}$ be the image measure on $(I,\mathcal{P}I)$. Fremlin breaks this down into two cases.

a) $\nu$ is atomless. As with any finite atomless measure space, there will be a measure preserving map $g\colon I\to[0,\gamma]$ for some $\gamma\ge0$, with respect to the Lebesgue measure $\lambda$ on $[0,\gamma]$. Using compactness, it can be shown that the sets on which $\lambda$ and $\nu\circ g^{-1}$ are well-defined coincide (precisely, $\mu$ is compact, so it is perfect, so $\nu\circ g^{-1}$ is perfect and therefore is Radon). The existence of non-Lebesgue sets will then give a contradiction unless $\gamma=0$, so $\mu(X)=0$.

b) $\nu$ has an atom $M\subseteq I$: In this case, $\mathcal{F}=\{F\subseteq M\colon\nu(M\setminus F)=0\}$ is a non-principal ultrafilter on $M$ which is closed under countable intersections. Again making use of compactness of $\mu$, this can be used to derive a contradition, but it requires some tricky set theory. I refer you to Fremlin (451P) for the full details of this argument.

**Update:** I will, however, give a brief overview of the ideas involved in (b). It is possible to reduce the problem to the case where $M$ is a regular uncountable ordinal and $\mathcal{F}$ is a *normal* ultrafilter. Using $[S]^n$ to denote the collection of size-$n$ subsets of a set $S$ and $[S]^{ < \omega}=\bigcup_{n=0}^\infty[S]^n$ for the collection of finite subsets, normal ultrafilters have the following property.

If $\mathcal{S}\subseteq[M]^{ < \omega}$ then there exists an $F\in\mathcal{F}$ such that, for each $n\ge0$, $[F]^n$ is either a subset of $\mathcal{S}$ or disjoint from $\mathcal{S}$.

See, Frelim (4A1L). This contradicts compactness as follows. Set $G_i=\bigcup\{E_j\colon j\in M, j\ge i\}$. Then choose $K_i\in\mathcal{K}$ with $K_i\subseteq G_i$ and $\mu(K_i) > 0$. Let $\mathcal{S}$ consist of the finite subsets $S\subset M$ such that $\bigcap_{i\in S}K_i=\emptyset$. Choose $F\in\mathcal{F}$ as above. It is not possible for $[F]^n$ to be a subset of $\mathcal{S}$. Otherwise, every $x\in X$ would be contained in no more than $n$ of the sets $\mathcal{K}^\prime=\{K_i\colon i\in F\}$. So, $\sum_{i\in F}\mu(K_i)\le n\nu(M)$. But, as this sum is over an uncountably infinite set of positive numbers, it should be infinite. Therefore, $[F]^{ < \omega}\cap\mathcal{S}=\emptyset$, and $\mathcal{K}^\prime$ satisfies the finite intersection property. So, by compactness, $\bigcap_{i\in F}G_i\supseteq\bigcap\mathcal{K}^\prime\not=\emptyset$. This contradicts the fact that, as $F\in\mathcal{F}$ is an unbounded subset of $M$, this intersection is empty.