How to test for and remove lead(II) sulfate from a sample of 1 M sulfuric acid?
To answer your question on how to test for Lead sulfate, I refer you to this paper, to quote:


  In 1929 Fischer [1] introduced a sensitive reagent for small quantities of lead which has found interesting applications in widely diverse fields. Diphenylthiocarbazone, or dithizone as it is commonly called...


And further:


  We have made no study of this reagent, except to find means of using it in our problem of measuring the solubility of lead sulfate. Dissolved in chloroform, dithizone imparts a noticeable green tint to the solution, provided the solution is not too concentrated. The lead-dithizone complex, however, has a bright cherry-red color.


Also of import;


  Dithizone is known to react with metals other than lead. The precautions taken to avoid interference by other metals are discussed on page 58. These precautions, together with the spectroscopic analysis of the lead salt and the redistillation of water and acid, make it highly improbable that the results were affected by other metals. Confirmation of this is found in the analysis of solutions of known lead content.  


If Pb is present, dilution appears to be the best path due to solubility.

[EDIT] Here is an theoretical path to the removal of heavy metals that does not require dilution. I am suggesting UV light therapy employing TiO2 photocatalyst to act on cold H2SO4 containing heavy metal impurities. Heavy metals are expected to precipitate.

With respect to the mechanics of the photolysis, I referred, for example, to this source. Per cited Equation [R1]:

$\ce{ hv -&gt; e- + h+ }$

Per Equation [R3], expected reaction with an electron hole:

$\ce{H2O + h+ -&gt; .OH + H+ }$

$\ce{H2SO4 = H+ + HSO4- }$

$\ce{H+ + e- = .H }$

where the hydrogen radical behaves as a e-/H+ pair, see Hydrometallurgy 2008: Proceedings of the Sixth International Symposium, p. 818, a commercial reductive leaching equation, to quote:


  $\ce{ PbS + 2 •H = Pb + H2S }$ (5) 


In the current case:

$\ce{ .H + PbSO4 -&gt; Pb (s) + HSO4- }$

However, the associated transient sulfate radical is also created:

$\ce{ .OH + HSO4- -&gt; H2O + .SO4- }$

Leading to the persulfate ion:

$\ce{ .SO4- + .SO4- -&gt; S2O8(2-) }$

So, the sulfate ions is not alone. Possibly more advantageous, further  pump air/oxygen into the solution during the UV light treatment. Then per Equation [R3], expect the superoxide radical anion formation:

$\ce{O2 + e- = .O2- }$

which becomes, in acidic conditions, the $\ce{.HO2}$ radical, which may similarly react with Lead sulfate as follows:

$\ce{ .HO2 + PbSO4 -&gt; PbO2 (s) + HSO4- }$

or, just oxidize any liberated Pb.

I would even suggest you try this process out by creating a test solution (with at least added Lead) and heat in H2SO4 to form a heavy metal impurity and then apply the UV light/TiO2 (and also added oxygen) therapy.