FFT usually requires power-of-2-sized windows, so let's say just DFT (alternatively use $1024$ samples).
A sine wave with a frequency of $6\:\mathrm{Hz}$ is not orthogonal to any of the $0\:\mathrm{Hz}$, $10\:\mathrm{Hz}$, ... waves with respect to the $L^2$ scalar product over an interval of $100\:\mathrm{ms}$, so it would in fact appear in all of the bins. That's the problem with a rectangular window function: unless you happen to start with a perfect superposition of only the quantized frequency values, you will end up with a horrible smear across the whole frequency range. To avoid having to introduce a nontrivial window function here, let's talk about compactly supported signals, like wavelets, centered in our DFT window. As you certainly know, such functions always have an intrinsic frequency indeterminacy, which essentially means that the (infinite) Fourier transform consists not of sharply defined (dirac) peaks but of Gaussian-like bell peaks. If the time confinedness was $100\: \mathrm{ms}$, the frequency indeterminacy will be more than $\frac1{100\:\mathrm{ms}}=10\:\mathrm{Hz}$. So as you see, the width of the DFT bins is not just a technical issue with the specific Fourier transform algorithm, it represents the general inability to define the frequency of a "correctly processable" signal more precisely than the bin width.
You probably know this already. Anyway, let's have a look now at a wavelet with a frequency centered about $60\:\mathrm{Hz}$, like you would get when window-functioning mains hum. Assuming the freq indeterminacy gets no bigger than necessary, this will give you a pretty sharp peak in the $55$ to $65\:\mathrm{Hz}$ bin, with only small values in the neighbouring bins - so we can approximate the total energy, that is, the $L^2$ norm of our signal, by just the square of the value in this bin (that's due to Bessel's equality). Likewise, if you were interested in the energy between $45$ and $105\:\mathrm{Hz}$, you would just sum up the squared values of those bins and get, correctly, the total energy of the wavelet. Where it gets interesting is when you want to know about the energy in the range $61$ to $63\:\mathrm{Hz}$. According to your proposal, this should be calculated as $\tfrac15$ of the squared value in the $55$ to $65\:\mathrm{Hz}$ bin, that is, $\tfrac15$th of the total energy of our wavelet. And that's pretty good actually, because as we said the energy of this wavelet is actually smeared over an interval of $10\:\mathrm{Hz}$ about $60\:\mathrm{Hz}$, so it's quite a reasonable approximation to say $\tfrac15$th of it is in the range $61$ to $63\:\mathrm{Hz}$!
What about a wavelet centered about $65\:\mathrm{Hz}$? If we DFT this, it will appear in both the $55$ to $65\:\mathrm{Hz}$ and $65$ to $75\:\mathrm{Hz}$ bins, with each values of $\sqrt{\tfrac12}$ of the total amplitude. If you now calculate the energy between $45$ and $105\:\mathrm{Hz}$, you will get
$$
0 + \sqrt{\tfrac12}^2E + \sqrt{\tfrac12}^2E + 0 + 0 = E
$$
so that's again the total energy, correctly. If you want the energy between $55$ and $65\:\mathrm{Hz}$, you get
$$
\sqrt{\tfrac12}^2E = \frac{E}2
$$
which is pretty reasonable, because in fact only about half of the signal energy lies in this band.
But where you start getting weird results is when you calculate the energy between $55$ to $56\:\mathrm{Hz}$, which results in
$$
\frac1{10}\sqrt{\tfrac12}^2E = \frac{E}{20}
$$
and compare it with the energy between $65$ to $66\:\mathrm{Hz}$, for which you would obviously get the same result. But then, the actual wavelet does not really have any notable frequency component at $55$ to $56\:\mathrm{Hz}$ at all, while $65$ to $66\:\mathrm{Hz}$ is where it is strongest!
In conclusion: it does make sense to do this interpolation, but it should be handled with care.
As I just notice, what you do is in fact not a linear interpolation, but just a $0$th order domain extension. A linear or higher-order interpolation would suffer less from the problem I just explained.
Your question doesn't contain enough detail to answer it. For example, you just just say you tried it and it didn't work. What exactly did you try? An FFT of your entire data set, or did you have to break it into chunks? If you broke it into chunks because it was a large set, there are many potential problems there and I can't begin to guess, but the correct procedure is complex. If you just did it in one go, did you pad your data? And so on.
That said, let me do my best to help you out with a few points:
it is usually a good idea to do this kind of filtering in the time or spacial domain, rather than the frequency domain. Why? Some of that is covered here: http://blog.bjornroche.com/2012/08/why-eq-is-done-in-time-domain.html
AnonSubmitter85 is not quite right in his comment when he says, "If you want to remove all frequencies above 3(Hz), then zeroing out the frequency bins that correspond to frequency values of 3(Hz) and higher will do that. That's just true by definition." What actually happens is a bit more complex: you guarantee that the signal at the frequency bins above 3Hz will be zero, but it says nothing about the values between the bins. In fact, a filter designed this way will have attenuation above 3Hz that fluctuates greatly depending on the exact frequency. This is known as filter ripple, and can be reduced in various ways, but not eliminated. Depending on how many bins and what your data looks like exactly, the effect of ripple may be little to no attenuation of high frequencies.
Time domain filters called IIR filters might "smear" the time response of your data, but it depends on the type and exactly what you are concerned about. If you are eliminating high frequencies to smooth out noise you are probably fine. Here is a blog post I wrote about basic audio filters that can guide you as well: http://blog.bjornroche.com/2012/08/basic-audio-eqs.html That post describes a class of IIR filters. If you are extremely concerned, you can filter your data twice: once regularly and then again with your data set reversed.
Best Answer
Well, from a practical perspective, a bandpass filter (or at least the devices I've worked with so far) doesn't exactly 'remove' the frequencies, even though it looks like that; it just attenuates them enough so that your system works as if they weren't there. You take the cutoff frequencies into account at design time to create a filter that suits your needs.
Now, from a more mathematical perspective, you can actually transform your signal, remove any component outside the frequency range you specify, and then get the resulting signal back to use it wherever you need it.