That's a great question! Indeed, in quantum mechanics, energy is only defined in expectation, and conservation of energy refers solely to expectation values. It's straightforward to show from the Schrodinger equation that this expectation value is constant in time.
However, measurement does not break conservation of energy as long as we keep track of the energy of the measurement apparatus. This has to be true because we can always consider the system plus the measurement apparatus as one joint quantum system, whose energy we know is conserved.
As a specific example, suppose we measure the position of a low-energy particle by firing a photon at it. If we measure the position very accurately, then the wavefunction collapses to a sharp spike, which has a very high energy which wasn't present in the original particle. This energy must have been absorbed from the incoming photon -- and indeed, by the de Broglie formula, $\lambda = h/p$, a high-precision position measurement can only be done using a high energy photon.
Some subtleties arise if the measurement device is macroscopic. For example, let's say that the photon in the previous example has a chance of missing, so the quantum state is a superposition
$$|\text{low energy particle and high energy photon} \rangle + |\text{high energy particle and low energy photon} \rangle.$$
The (expectation value) of the energy is the same as before at this point. Now suppose we use a high-energy photon detector and read off the result in the lab, giving a superposition of
$$|\text{low energy particle, you see '1' on detector} \rangle + |\text{high energy particle, you see '0' on detector} \rangle.$$
At this point energy is still perfectly conserved if you account for the energy of everything in each branch, but if you subscribe to the Copenhagen interpretation, you might say this state is unacceptable because macroscopic observers can't be in superpositions. In other words, you say the state must be
$$|\text{low energy particle, you see '1' on detector} \rangle \text{ or } |\text{high energy particle, you see '0' on detector} \rangle$$
rather than a superposition.
This is also fine. The problem is that one is then tempted to remove oneself and the detector from the picture entirely, arriving at
$$|\text{low energy particle} \rangle \text{ or } |\text{high energy particle} \rangle$$
which are two states with different energy. This thinking led the pioneers of quantum mechanics to propose that energy was conserved only on average (e.g. in BKS theory, see here).
This is how it's usually presented in introductory textbooks, but that is only done for simplicity. When you think about it, it's a profoundly unphysical picture -- it basically treats measurements as occurring magically, without any physical cause. Modern-day users of the Copenhagen interpretation don't use it in this unphysical way, and we have known for almost a century that energy non-conserving theories like BKS are completely wrong. Energy is exactly conserved in all genuine interpretations of quantum mechanics.
